(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 156676, 5576]*) (*NotebookOutlinePosition[ 157439, 5603]*) (* CellTagsIndexPosition[ 157395, 5599]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["CONIC SECTIONS", "Title", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["Read Me! (how to use this program)", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "This notebook is written using ", StyleBox["Mathematica", FontSlant->"Italic"], ", a highly sophisticated mathematical software program capable of doing \ intricate mathematics", ". ", "You will need to know only a few ", StyleBox["very basic", FontVariations->{"Underline"->True}], " things about ", StyleBox["Mathematica", FontSlant->"Italic"], " to be able to use this notebook; the programming has been done for you." }], "Text"], Cell[TextData[{ "\t1. ", StyleBox["Opening a section of the notebook", FontWeight->"Bold"], ": Scroll down the screen using the mouse on the scrollbar at the right of \ this screen until you see what appears to be a table of contents. Directly \ to the right of each topic is a short blue bracket sign. Some brackets have \ a small arrow or triangle at the bottom. This arrow indicates that there is \ hidden text which can be viewed by clicking with the mouse on the bracket \ containing the arrow. Try this on one of the arrowed brackets below. When \ you click on the bracket, new text should appear. When you are finished, you \ can close that section of the notebook by double-clicking on the same (now \ longer) bracket. If you are unsure which one it is, scroll up to the \ beginning of the section; it is the one that ", StyleBox["begins", FontSlant->"Italic"], " there", ". ", "Try closing the section you just opened." }], "Text"], Cell[TextData[{ StyleBox["\t", FontWeight->"Bold"], "2. ", StyleBox["Activating a program command", FontWeight->"Bold"], ": The ", StyleBox["Mathematica", FontSlant->"Italic"], " programming always appears in bold type on a yellow background. To \ activate a program, use the mouse to move the \"I\" shaped cursor inside the \ yellow box and click once so that the straight line \"|\" cursor appears \ anywhere inside the yellow box. Now depress the \"shift\" and \"return\" \ keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], ". Try this with the following short program which generates a random \ number between 1 and 10", ". ", "Remember to click on these two keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], "." }], "Text"], Cell[BoxData[ \(Random[Integer, {1, 10}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\t3. ", StyleBox["Changing a program:", FontWeight->"Bold"], " Generally, you should not change any type in the yellow boxes. However, \ a few of the ", StyleBox["Mathematica", FontSlant->"Italic"], " programs have been designed so that you can change parts of the program. \ The parts that you should change are displayed in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], ". To change these parts, simply highlight the ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " type by sweeping over it with the cursor, and type in your change. Read, \ and take seriously, any instruction saying ", StyleBox["*do not change anything below this line*", FontColor->RGBColor[1, 0, 0]], ". Changing something there could radically change the commands, causing ", StyleBox["Mathematica", FontSlant->"Italic"], " to 'beep' in an error protest--or give you an answer to a totally \ different question!\n\n\t4. ", StyleBox["Quitting the program", FontWeight->"Bold"], ": When you have finished working and want to quit, click in the small box \ at the upper lefthand corner of the window to close it. Then choose `quit' \ or `exit' from the File menu. ", StyleBox["Mathematica", FontSlant->"Italic"], " will ask you if you want to save your changes. Say NO! Otherwise you \ will have a different notebook from the one you downloaded, and any \ misinformation you might have entered into the notebook will persist", ". ", "If for any reason, you need to start with a fresh notebook in its original \ form, you can always trash the one you've been working with and download a \ new one off the Web." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Introduction", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["\tConics", FontWeight->"Bold", FontVariations->{"Underline"->True}], ", or conic sections, are curves obtained by the intersection of a plane \ and a right circular cone. Below we see a picture of such a cone. It is \ formed by rotating a line ", StyleBox["g", FontSlant->"Italic"], " about an axis (the z-axis in the picture below) at a fixed angle, where ", StyleBox["g", FontSlant->"Italic"], " meets the axis in a point V, the ", StyleBox["vertex", FontSlant->"Italic"], " of the cone", ". ", "The line g is called the ", StyleBox["generator", FontSlant->"Italic"], " of the cone." }], "Text"], Cell[BoxData[ \(\(cone\ = \ ParametricPlot3D[{u\ Sin[v], u\ Cos[v], \ u}, {u, \(-7\), 7}, {v, 0, 2 \[Pi]}, Axes -> False, Boxed -> False, ViewPoint -> {\(-2\), 2, 1.5}]; \)\)], "Input", FontColor->GrayLevel[0.500008], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tFirst try to visualize a curve obtained by slicing through the cone with \ a plane perpendicular to the axis. Then begin to tilt the plane slightly to \ one just meeting the upper half or lower half of the cone, then to one \ parallel to the generator, and finally to one tilted so much it slices \ through both halves. Let none of the planes pass through the vertex of the \ cone. What do the various slices look like? These curves represent the \ various conic sections: ", StyleBox["circle", FontWeight->"Bold"], ", ", StyleBox["ellipse", FontWeight->"Bold"], ", ", StyleBox["parabola", FontWeight->"Bold"], " and ", StyleBox["hyperbola", FontWeight->"Bold"], ". ", "Draw your own pictures, then compare with the ones below." }], "Text", FontSize->13], Cell[BoxData[ \(p3\ = \ ParametricPlot3D[{x, y, x + 2}, {x, \(-5\), 5}, {y, \(-5\), 5}, Axes -> False, Boxed -> False, DisplayFunction -> Identity]; Show[{cone, p3}, DisplayFunction -> $DisplayFunction, ViewPoint -> {\(-2\), \(-2\), 1.5}]; Show[{cone, p3}, DisplayFunction -> $DisplayFunction, ViewPoint -> {0, \(-2\), 5}]; \n Print["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[BoxData[ \(p1\ = \ ParametricPlot3D[{x, y, 1}, {x, \(-5\), 5}, {y, \(-5\), 5}, Axes -> False, Boxed -> False, DisplayFunction -> Identity]; Show[{cone, p1}, DisplayFunction -> $DisplayFunction, ViewPoint -> {\(-2\), \(-2\), 1.5}]; Show[{cone, p1}, DisplayFunction -> $DisplayFunction, ViewPoint -> {0, 0, 5}]; \nPrint["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[BoxData[ \(p2\ = \ ParametricPlot3D[{x, y, \(2\ x\)\/3 + 5\/3}, {x, \(-5\), 5}, {y, \(-5\), 5}, Axes -> False, Boxed -> False, DisplayFunction -> Identity]; Show[{cone, p2}, DisplayFunction -> $DisplayFunction, ViewPoint -> {\(-2\), \(-2\), 1.5}]; Show[{cone, p2}, DisplayFunction -> $DisplayFunction, ViewPoint -> {0, 0, 5}]; \nPrint["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[BoxData[ \(p4\ = \ ParametricPlot3D[{x, y, 3\ x + 4}, {x, \(-4\), 1.5}, {y, \(-7\), 7}, Axes -> False, Boxed -> False, DisplayFunction -> Identity]; Show[{cone, p4}, DisplayFunction -> $DisplayFunction, ViewPoint -> {\(-2\), \(-2\), 1.5}]; Show[{cone, p4}, DisplayFunction -> $DisplayFunction, ViewPoint -> {\(-2\), 0, 3}]; \n Print["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ In the following sections, we take a closer look at each of these types of \ curves, taking a much different perspective in defing the conic sections.\ \>", "Text", FontSize->13] }, Closed]], Cell[CellGroupData[{ Cell["Section 1: The Parabola", "Section", FontSize->14, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["1. A parabola with vertex at the origin", "Subsection", Background->None], Cell[TextData[{ "As seen above, a parabola is obtained by slicing through a cone with a \ plane parallel to a generator, but not passing through a vertex. However, we \ can also take a different approach in defining it. \n\tFix a line D and a \ point F not on D in the plane. A ", StyleBox["parabola", FontWeight->"Bold"], " is the collection of points P in the plane so that the distance from D to \ P equals the distance from F to P, i.e. d(P, D) ", "= ", "d(P, F). The line D is called the ", StyleBox["directrix", FontWeight->"Bold"], " and the point P the ", StyleBox["focus", FontWeight->"Bold"], ". Further, the line through F perpendicular to D is the ", StyleBox["axis of symmetry", FontWeight->"Bold"], ", and this axis intersects the parabola at the ", StyleBox["vertex", FontWeight->"Bold"], " of the parabola." }], "Text", FontSize->13], Cell[BoxData[ \(Plot[y\^2\/4, {y, \(-4\), 4}, Axes -> False, PlotRange -> {\(-3\), 5}, PlotStyle -> {Thickness[0.01], Hue[0.9]}, AspectRatio -> 1, Epilog -> {AbsolutePointSize[6], Hue[0.6], Point[{0, 1}], \ Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], { 0, 1.3}], Line[{{0, 1}, {2, 1}}], Line[{{2, 1}, {2, \(-1\)}}], Line[{{0, 1}, {0, 0}}], Line[{{0, 0}, {0, \(-1\)}}], Line[{{0, 1}, {3, 9/4}}], Line[{{0, 1}, {1, 1/4}}], Line[{{1, 1/4}, {1, \(-1\)}}], Line[{{3, 9/4}, {3, \(-1\)}}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], { 3.3, 9/4}], Line[{{\(-4\), \(-1\)}, {4, \(-1\)}}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], { \(-4\), \(-0.8\)}]}]; \n Print["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ \tLet's look at a particularly simple case, where the axis of symmetry is the \ x-axis, and the vertex V is at the origin. Since V is on the parabola, it is \ equidistant from the directrix and the focus. So say F = (a, 0) and the D is \ the vertical line x = -a, with a > 0.\ \>", "Text", FontSize->13], Cell[BoxData[ \(\(Plot[{2 \@ x, \(-2\) \@x}, {x, 0, 3}, PlotRange -> {{\(-2\), 3}, {\(-4\), 4}}, PlotStyle -> {{Thickness[0.01], Hue[0.9]}, {Thickness[0.01], Hue[0.9]}}, Ticks -> None, AxesLabel -> {x, y}, AspectRatio -> Automatic, Epilog -> {AbsolutePointSize[6], Dashing[{0.005, 0.02}], Point[{1, 0}], Line[{{1, 0}, {2, 2 \@ 2}}], Line[{{2, 2 \@ 2}, {\(-1\), 2 \@ 2}}], Line[{{\(-1\), \(-4\)}, {\(-1\), 4}}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], { 2, 3}], Text[ StyleForm["\", FontSize -> 14, FontWeight -> "\"], { \(-1.2\), \(-3.5\)}]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Then any point P = (x, y) on the parabola satisfies d(P, D) = d(P, F), \ i.e.\n\n\t", Cell[BoxData[ \(TraditionalForm\`\@\(\((x - a)\)\^2 + y\^2\)\)]], " = |x-(-a)| using the distance formula,\n\t\n\tor \n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\((x - a)\)\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\((x + a)\)\^2\)]], " by squaring both sides.\n\t\n\tExpanding and simplifying gives ", Cell[BoxData[ \(TraditionalForm\`\(y\^2\ \)\)]], " = 4ax, the equation of a parabola with vertex at the origin, focus at \ (a,0), and directrix x ", "= ", "-a, a > 0.\n\tThe constant 4a in the above equation has a simple geometric \ interpretation. Take a line perpendicular to the axis of symmetry and \ passing through the focus. This line meets the parabola in two points. The \ distance between these two points on the curve is 4a and is called the ", StyleBox["latus rectum", FontWeight->"Bold"], "." }], "Text", FontSize->13], Cell[BoxData[ \(Plot[{\@\(4\ x\), \(-\@\(4\ x\)\)}, {x, 0, 2}, PlotStyle -> {{Thickness[0.01], Hue[0.9]}, {Thickness[0.01], Hue[0.9]}}, PlotRange -> {{\(-2\), 2}, {\(-2\), 2}}, AspectRatio -> 1, Epilog -> {AbsolutePointSize[6], Hue[0.6], Point[{1, 2}], Point[{1, \(-2\)}], Line[{{1, 2}, {1, \(-2\)}}], Text[StyleForm["\<4a\>", FontSize -> 14, FontWeight -> "\"], {1.2, 0.2}]}]; \n Print["\"]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the equation of a parabola with vertex at (0,0) and focus at (5,0)", ". ", "Then sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n Plot[{\@\(20\ x\), \(-\@\(20\ x\)\)}, {x, 0, 5}, PlotStyle -> Thickness[0.01]]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the vertex, focus, and directrix of the parabola with equation ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = 12x. Then sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n Plot[{\@\(12\ x\), \(-\@\(12\ x\)\)}, {x, 0, 5}, PlotStyle -> Thickness[0.01]]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox[" ", FontWeight->"Bold"], "What changes if the focus is to the ", StyleBox["left", FontSlant->"Italic"], " of the vertex instead of the right?", StyleBox[" \n\nExercise", FontWeight->"Bold"], ": Find the general equation of a parabola with vertex at the origin, focus \ at (-a, 0), and directrix x ", "= ", "a, where a > 0." }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ "Print", "[", "\"\ 0.\>\"", "]"}]], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNotice that the curve in this case is just the reflection across the \ y-axis of the first case. So we could also have found its equation by \ substituting -x for x to obtain the equation ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = -4ax.\n\tNow use the following program to practice these graphs. You \ may enter the value of a. If a is positive, you'll have the first case \ above; if negative, the second." }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[a]\), ";", "\n", StyleBox[\(a\ = \ \(-2\)\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(Plot[{\@Abs[4\ a\ x], \(-\@Abs[4 a\ x]\)}, {x, 0, a}, PlotStyle -> {{Thickness[0.01], Hue[0.9]}, {Thickness[0.01], Hue[0.9]}}, PlotRange -> {{\(-2\)\ a, 2\ a}, {\(-2\)\ a, 2\ a}}, AspectRatio -> 1]\), ";"}], "\n", "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox[\( (*shift - enter*) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Next we'll keep the vertex at the origin, but take the axis of symmetry to \ be the y-axis instead of the x-axis. Duplicate the above computations to \ find the vertex and directrix and the resulting equation. You'll have two \ cases, depending on whether the focus is above or below the vertex", ". ", StyleBox["Try this on your own", FontVariations->{"Underline"->True}], " before going to the summary below." }], "Text", FontSize->13], Cell[TextData[{ StyleBox[ "Summary:\n Equation of a parabola with vertex at the origin and focus on \ the x or y-axis\n\nvertex\t\tfocus\t\tdirectrix\t\tequation", FontSize->14, FontWeight->"Bold"], StyleBox["\n", FontSize->14], StyleBox["\n(0, 0)\t\t(a, 0)\t\tx", FontSize->14, FontWeight->"Bold"], " = ", StyleBox["-a\t\t\t", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["4ax\n(0, 0)\t\t(-a, 0)\t\tx", FontSize->14, FontWeight->"Bold"], " = ", StyleBox["a\t\t\t", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["-4ax\n(0, 0)\t\t(0, a)\t\ty", FontSize->14, FontWeight->"Bold"], " = ", StyleBox["-a\t\t\t", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["4ay\n(0, 0)\t\t(0, -a)\t\ty", FontSize->14, FontWeight->"Bold"], " = ", StyleBox["a\t\t\t", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSize->14, FontWeight->"Bold"], " ", "= ", StyleBox["-4ay", FontSize->14, FontWeight->"Bold"], StyleBox["\n", FontSize->14], StyleBox["\n", FontSize->14, FontWeight->"Bold"], "Notice that all these curves are symmetric with respect to the axis of \ symmetry." }], "Text", FontSize->14], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ " Find the equation of the parabola with vertex at the origin, axis of \ symmetry the y-axis and containing the point (4, 1)", FontSlant->"Plain"], ". ", StyleBox["Graph the curve.", FontSlant->"Plain"] }], "Text", FontSize->13, FontSlant->"Italic"], Cell[BoxData[ \(Print[ \*"\"\\""]; \n Plot[x\^2/16, {x, \(-4\), 4}, PlotStyle -> {Thickness[0.01], Hue[0.6]}, PlotRange -> {{\(-4\), 4}, {\(-1\), 2}}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["2. A parabola with vertex at (h, k)", "Subsection", Background->None], Cell[TextData[{ StyleBox["\tLet's now generalize a bit", FontSize->12], ". ", StyleBox[ "We'll let the vertex of the parabola be at any point (h, k), but we'll \ still ask that the axis of symmetry be vertical or horizontal", FontSize->12], ". ", StyleBox[ "(Note: in the last section of this notebook, we'll tilt the axes too!) \ The graph is then simply a translation of a parabola with vertex at the \ origin--shifted horizontally h units and vertically k units", FontSize->12], ". ", StyleBox[ "The parabola shown below is shifted to the right 2 units and up 3 from \ original blue parabola with vertex at the origin.", FontSize->12] }], "Text", FontSize->9], Cell[BoxData[ \(\(Plot[{1\/4\ x\^2, \ \(1\/4\) \((x - 2)\)\^2 + 3}, {x, \(-5\), 7}, PlotStyle -> {{Hue[0.3], Hue[0.6]}, {Thickness[0.01], Thickness[0.01]}}, Epilog -> {AbsolutePointSize[6], Point[{0, 0}], Point[{2, 3}]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ We can obtain the equations of such parabolas by replacing x by x - h and y \ by y - k in the table above to obtain the following:\ \>", "Text"], Cell[TextData[{ StyleBox[ "Equation of a parabola with vertex at (h, k) and axis of symmetry vertical \ or horizontal.\n\nvertex\t\tfocus\t\tdirectrix\t\tequation", FontWeight->"Bold"], "\n", StyleBox["\n(h, k)\t\t(h + a, k)\tx", FontWeight->"Bold"], " = ", StyleBox["-a + h\t\t", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\((y - k)\)\^2\)], FontWeight->"Bold"], " = ", StyleBox["4a(x - h)\n(h, k)\t\t(h - a, k)\tx", FontWeight->"Bold"], " = ", StyleBox["a + h\t\t", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\((y - k)\)\^2\)], FontWeight->"Bold"], " = ", StyleBox["-4a(x - h)\n(h, k)\t\t(0, h + a)\ty", FontWeight->"Bold"], " = ", StyleBox["-a + h\t\t", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\(\((x - h)\)\^2\ \)\)], FontWeight->"Bold"], " = ", StyleBox["4a(y - k)\n(h, k)\t\t(0, h - a)\ty", FontWeight->"Bold"], " = ", StyleBox["a + h\t\t", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\((x - h)\)\^2\)], FontWeight->"Bold"], " ", "= ", StyleBox["-4a(y - k)", FontWeight->"Bold"], "\n", StyleBox[ "\nNotice that all these curves are symmetric with respect to the axis of \ symmetry.", FontWeight->"Bold"] }], "Text", FontSize->14], Cell[TextData[{ StyleBox["Try some exercises:", FontWeight->"Bold", FontVariations->{"Underline"->True}], "\n\n", StyleBox["Ex1:", FontWeight->"Bold"], " Find the equation of the parabola with vertex at (-2, 3) and focus at (0, \ 3)", ". ", "Graph it by hand. " }], "Text", FontSize->13], Cell[TextData[{ StyleBox["Ex2: ", FontWeight->"Bold"], "Find the vertex, focus, and directrix of each parabola. \n\t\t\n\t\ta) \ x", Cell[BoxData[ \(TraditionalForm\`\^2\)]], " + 8x = 4y - 8\t\t(b) ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - 4y + 4x + 4 ", "= ", "0" }], "Text", FontSize->13], Cell[TextData[{ StyleBox["Ex3: ", FontWeight->"Bold"], "Write the equation for the following parabola.\n", Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1.1 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations 0.1 0.1 0.2 0.1 [ [.3 .1875 -3 -9 ] [.3 .1875 3 0 ] [.5 .1875 -3 -9 ] [.5 .1875 3 0 ] [.7 .1875 -3 -9 ] [.7 .1875 3 0 ] [.9 .1875 -3 -9 ] [.9 .1875 3 0 ] [.0875 0 -12 -4.5 ] [.0875 0 0 4.5 ] [.0875 .4 -6 -4.5 ] [.0875 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Cell["Ex2:", "Text", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(Print[ \*"\"\<(a) Complete the square and get (x - 4\!\(\()\^2\)\) = 4(y + \ 2). The \nvertex is (4, -2), the focus is (4, -1), and the directrix is y = \ -3.\>\""]\ ; \n Print[\*"\"\<(b) Complete the square and obtain (y - 2\!\(\()\^2\)\) = \ -4x, so that the vertex is (0, 2), the focus is (-1, 2), and the directrix is \ x = 1.\>\""]\)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell["Ex3:", "Text", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(Print[\*"\"\\""]\)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Section 2: The Ellipse", "Section", FontSize->14, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[StyleBox["CLICK IN HERE FIRST! ", FontVariations->{"Underline"->True}]], "Text", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(Needs["\"]\)], "Input", FontColor->RGBColor[1, 0, 0], Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell["1. An ellipse with center at the origin", "Subsection"], Cell[TextData[{ "\tFix two points, ", Cell[BoxData[ \(TraditionalForm\`F\_1\ and\ F\_2, \ in\ the\ plane\)]], ". An ", StyleBox["ellipse", FontWeight->"Bold"], " is the set of all points in the plane so that the sum of the distances \ from the two fixed points is constant. The points ", Cell[BoxData[ \(TraditionalForm\`F\_1\ and\ F\_2\)]], " are called the ", StyleBox["foci", FontWeight->"Bold"], " of the ellipse. In other words, P is on the ellipse if and only if d(", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], ",P) + d(", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ",P) = 2a for some constant a. (Note the '2a' is for convenience only, as \ we'll see shortly.)" }], "Text", FontSize->13], Cell[BoxData[ \(\(ImplicitPlot[x\^2\/9 + y\^2\/4 - 1\ == \ 0, {x, \(-3\), 3}, PlotStyle -> {Hue[0.6], \ Thickness[0.01]}, Epilog -> {AbsolutePointSize[6], Hue[0.7], Thickness[0.001], Point[{\(-\@5\), 0}], Point[{\@5, 0}], Line[{{\(-\@5\), 0}, {1, \@32\/3}}], Line[{{1, \@32\/3}, {\@5, 0}}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], { 1, 1.6}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {\(-2.2\), 0.2}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {2.4, 0.2}]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tThe axis along the longer direction of the ellipse and containing the \ foci is the ", StyleBox["major axis", FontWeight->"Bold"], "; the midpoint of the foci is the ", StyleBox["center", FontWeight->"Bold"], "; and the line through the center perpendicular to the major axis is the ", StyleBox["minor axis", FontWeight->"Bold"], ". Note that the ellipse is symmetric with respect to the major and minor \ axes. There are two ", StyleBox["vertices", FontWeight->"Bold"], " to an ellipse: the points where the major axis meets the ellipse.\n\t", StyleBox["Note", FontWeight->"Bold"], ": A ", StyleBox["circle", FontWeight->"Bold"], " is just a special case of an ellipse, where the two foci coincide. \ Letting F = ", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], " = ", Cell[BoxData[ \(TraditionalForm\`F\_\(\(2, \)\ \)\)]], " we see from the defintion of ellipse, d(", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], ",P) + d(", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ",P) = 2 d(F,P) is a constant--namely the diameter of the circle! From \ here on, we shall refer to a circle as this particular type of ellipse. In \ this case, the axes and vertices are ", StyleBox["not", FontSlant->"Italic"], " well-defined." }], "Text", FontSize->13], Cell[TextData[{ "\tNow we will develop the equation of an ellipse. We'll place its center \ at the origin and the foci at (-c,0) and (c,0). Let P = (x,y) be a point on \ the ellipse. Using d(", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], ",P) + d(", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ",P) = 2a and the distance formula, we have\n\n", Cell[BoxData[ \(TraditionalForm\`\@\(\((x - \((\(-c\))\))\)\^2 + y\^2\)\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\@\(\((x - c)\)\^2 + y\^2\)\)]], " = 2a.\n\nSquaring, simplifying, and substituting ", Cell[BoxData[ \(TraditionalForm\`b\^2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(a\^2 - c\^2\ \)\)]], "gives (eventually)\n\n", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)], FontSize->14, FontWeight->"Bold"], StyleBox["+", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["1", FontSize->14, FontWeight->"Bold"], StyleBox[",\n", FontSize->14], StyleBox["\n", FontSize->14, FontWeight->"Bold"], StyleBox[ "the equation of an ellipse with center at the origin and foci at (-c, 0) \ and (c, 0)", FontSize->12, FontWeight->"Bold"], StyleBox[".", FontSize->12], "\n\nNote that a > c, and that a \[GreaterEqual] b > 0. When a = b, we \ have a circle." }], "Text", FontSize->13], Cell[TextData[{ "A simple variation of the computations above shows that the equation of an \ ellipse with center at the origin but major axis along the y-axis is ", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)]], " = 1 where b > a > 0. Now", Cell[BoxData[ \(TraditionalForm\`\(\ \(a\^\(2\ \ \) = \)\ \)\)]], " ", Cell[BoxData[ \(TraditionalForm\`b\^2 - c\^2\)]], ". For example, the graph of ", Cell[BoxData[ \(TraditionalForm\`x\^2\/4\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\/9\)]], " = 1 is given below." }], "Text", FontSize->13], Cell[BoxData[ \(\(ImplicitPlot[x\^2\/4 + y\^2\/9 - 1\ == \ 0, {x, \(-2\), 2}, PlotStyle -> {Hue[0.6], \ Thickness[0.011]}, Epilog -> {AbsolutePointSize[6], Hue[0.9], Thickness[0.001], AbsolutePointSize[6], Point[{0, \(-\@5\)}], Point[{0, \@5}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {0.3, 2.2}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {0.3, \(-2.2\)}]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "So, if the number under x is larger, the ellipse is wider than it is tall; \ if the number under y is larger, it's taller than it is long. You do not \ have to memorize this. Just plot the points along the major and minor axes \ and it will be apparent!\n\tHere's a program to let you practice drawing \ ellipses with center at the origin. You set the values of a and b; it will \ graph the ellipse ", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)]], "+", Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)]], " = 1." }], "Text", FontSize->13], Cell[BoxData[ RowBox[{\(Clear[a, b]\), ";", "\n", RowBox[{ StyleBox["a", FontColor->GrayLevel[0]], StyleBox[" ", FontColor->GrayLevel[0]], StyleBox["=", FontColor->GrayLevel[0]], StyleBox[" ", FontColor->GrayLevel[0]], StyleBox["4", FontColor->RGBColor[0, 0, 1]]}], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], RowBox[{ StyleBox["b", FontColor->GrayLevel[0]], StyleBox[" ", FontColor->GrayLevel[0]], StyleBox["=", FontColor->GrayLevel[0]], StyleBox[" ", FontColor->GrayLevel[0]], StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(ImplicitPlot[x\^2\/a\^2 + y\^2\/b\^2 - 1\ == \ 0, {x, \(-a\), a}, PlotStyle -> {Hue[0.6], \ Thickness[0.01]}]\), ";"}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the points where ", Cell[BoxData[ \(TraditionalForm\`\(x\^2\ \ \)\/25\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`y\^2\/4\)]], " = 1 crosses the x and y axes. Use these 4 points only to sketch the \ graph. Which are the vertices? Find the foci." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[x\^2\/25 + y\^2\/4 - 1\ == \ 0, {x, \(-5\), 5}, PlotStyle -> {Hue[0.6], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": For the ellipse given by ", Cell[BoxData[ \(TraditionalForm\`x\^2\/9\)]], "+", Cell[BoxData[ \(TraditionalForm\`y\^2\/16\)]], " = 1, find the vertices, and draw the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[x\^2\/9 + y\^2\/16 - 1\ == \ 0, {x, \(-3\), 3}, PlotStyle -> {Hue[0.8], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the equation of the ellipse with center at the origin, one focus at \ (4,0), and a vertex at (6,0). Sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[x\^2\/36 + y\^2\/20 - 1\ == \ 0, {x, \(-6\), 6}, PlotStyle -> {Hue[0.8], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["2. An ellipse with center at (h,k)", "Subsection"], Cell[TextData[{ "To get the equation of an ellipse with center at a point (h, k) and major \ axis parallel to either the x or y-axis, we again translate. Shift \ horizontally h units and vertically k units; hence substitute x - h in place \ of x and y - k in place of y to get\n \n", StyleBox[" ", FontSize->14], Cell[BoxData[ \(TraditionalForm\`\((x - h)\)\^2\/a\^2\)], FontSize->14, FontWeight->"Bold"], StyleBox["+ ", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\((y - k)\)\^2\/b\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["1\n \n", FontSize->14, FontWeight->"Bold"], StyleBox["i) if a > b > 0, the major axis is horizontal", FontSize->12, FontWeight->"Bold"], StyleBox[". ", FontSize->12], StyleBox["The foci are (h+c, k) and (h-c, k), where ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`b\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`a\^2 - c\^2\)], FontWeight->"Bold"], ".", StyleBox[ "\nii) if b > a > 0, the major axis is vertical and the foci are (h, k+c) \ and (h, k-c), ", FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(a\^2\), FontWeight->"Bold"], TraditionalForm]]], StyleBox[" = ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`b\^2 - c\^2\)], FontWeight->"Bold"], "." }], "Text", FontSize->13], Cell["\<\ Use the program below to practice graphing off-center ellipses. You can \ choose the a, b, h, and k values.\ \>", "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[a, b, h, k]\), ";", "\n", StyleBox[\(a\ = \ 4\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(b\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(h\ = \ 5\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(k\ = \ 3\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(ImplicitPlot[ \((x - h)\)\^2\/a\^2 + \((y - k)\)\^2\/b\^2 - 1\ == \ 0, {x, h - a, h + a}, AxesOrigin -> {0, 0}, PlotRange -> {{\(-1\), 10}, {\(-1\), 6}}, PlotStyle -> {Hue[0.8], Thickness[0.01]}]\), ";"}], "\n", "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t \ ", StyleBox[\( (*shift - enter*) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the equation for the ellipse with center at (3, 2), one focus at \ (1, 2), and one vertex at (6, 2). Sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[ \((x - 3)\)\^2\/9 + \((y - 2)\)\^2\/5 - 1\ == \ 0, {x, \(-1\), 6}, AxesOrigin -> {0, 0}, PlotRange -> Automatic, PlotStyle -> {Hue[0.8], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the center, major axis, foci and vertices of each of the following. \ Then sketch the graph.\n\t\ti) 9(", Cell[BoxData[ \(TraditionalForm\`\(x - 3)\)\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`\((y\ - \ 2)\)\^2\)]], " = 36 (note: first divide to put it into standard form)\n\t\tii) 4", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " + 4y = 0 (note: you must complete the square to get it into standard \ form)" }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[ \((x - 3)\)\^2\/4 + \((y - 2)\)\^2\/36 - 1\ == \ 0, {x, \(-1\), 6}, AxesOrigin -> {0, 0}, PlotRange -> {{\(-1\), 6}, {\(-5\), 9}}, PlotStyle -> {Hue[0.6], Thickness[0.01]}, \ Epilog -> {AbsolutePointSize[6], Hue[0.9], Thickness[0.01], Point[{3, 2 + \ 4 \@ 2}], Point[{3, \ 2\ - \ 4 \@ 2}], Point[{3, 8}], Point[{3, \(-4\)}]}]; \n Print[\*"\"\\""]; \n ImplicitPlot[x\^2 + \((y + 2)\)\^2\/4 - 1\ == \ 0, {x, \(-5\), 5}, AxesOrigin -> {0, 0}, PlotRange -> {{\(-1.5\), 1.5}, {\(-5\), 1}}, PlotStyle -> {Hue[0.6], Thickness[0.01]}, \ Epilog -> {AbsolutePointSize[6], Hue[0.9], Thickness[0.01], Point[{0, \ \(-2\) + \@3}], Point[{0, \ \(-2\) - \@3}], Point[{0, \ 0}], Point[{0, \ \(-4\)}]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Section 3: The Hyperbola", "Section", FontSize->14, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["CLICK IN HERE FIRST!", FontVariations->{"Underline"->True}], " ", StyleBox["(Unless you've done it in the previous section.)", FontWeight->"Plain"] }], "Text", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(Needs["\"]\)], "Input", FontColor->RGBColor[1, 0, 0], Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell["1. A hyperbola with center at the origin", "Subsection"], Cell[TextData[{ "Once again, we begin with the defining equation. Fix two points, ", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ", in the plane. A ", StyleBox["hyperbola", FontWeight->"Bold"], " is the set of all points in the plane so that the difference of the \ distances from ", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], " is constant. The points ", Cell[BoxData[ \(TraditionalForm\`F\_1\ and\ F\_2\)]], " are called the ", StyleBox["foci", FontWeight->"Bold"], " of the hyperbola. So P is on the hyperbola if and only if d(", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], ",P) - d(", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ",P) = ", Cell[BoxData[ \(TraditionalForm\`\[PlusMinus]\)]], "2", StyleBox["a", FontSlant->"Italic"], " for some constant ", StyleBox["a", FontSlant->"Italic"], "." }], "Text", FontSize->13], Cell[BoxData[ \(\(ImplicitPlot[x\^2\/9 - y\^2\/4 - 1\ == \ 0, {x, \(-5\), 5}, Axes -> False, PlotStyle -> {Hue[0.9], Thickness[0.008]}, Epilog -> {AbsolutePointSize[6], Hue[0.6], Point[{\(-\@13\), 0}], Point[{\@13, 0}], Line[{{\(-5\), 0}, {5, 0}}], {Dashing[{0.05, .05}], Line[{{\(-\@13\), 0}, {4, \@28\/3}}], Line[{{4, \@28\/3}, {\@13, 0}}]}, Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {\(-3.5\), 0.3}], Text[StyleForm["\", FontSize -> 14, FontWeight -> "\"], {4.2, 0.3}]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "The line through the foci is called the ", StyleBox["transverse axis,", FontWeight->"Bold"], " and the midpoint of the foci is the ", StyleBox["center", FontWeight->"Bold"], " of the hyperbola. The ", StyleBox["vertices", FontWeight->"Bold"], " of the hyperbola are the points where the hyperbola meets the transverse \ axis.\n\tSuppose the transverse axis is the x-axis, the center is the origin, \ and the foci are ", Cell[BoxData[ \(TraditionalForm\`\(\(F\_1\ = \)\ \)\)]], "(-", StyleBox["c", FontSlant->"Italic"], ", 0) and ", Cell[BoxData[ \(TraditionalForm\`F\_\(2\ \)\)]], " = (", StyleBox["c", FontSlant->"Italic"], ", 0). Let's again use the definition to find the equation of such a \ hyperbola. Let ", StyleBox["P", FontSlant->"Italic"], " = (", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], ") be any point on the hyperbola. According to the definition, d(", Cell[BoxData[ \(TraditionalForm\`F\_1\)]], ",P) - d(", Cell[BoxData[ \(TraditionalForm\`F\_2\)]], ",P) = ", Cell[BoxData[ \(TraditionalForm\`\[PlusMinus]\)]], "2", StyleBox["a", FontSlant->"Italic"], ", i.e.\n\t\n", Cell[BoxData[ \(TraditionalForm\`\@\(\((x - \((\(-c\))\))\)\^2 + y\^2\)\)]], "- ", Cell[BoxData[ \(TraditionalForm\`\@\(\((x - c)\)\^2 + \ y\^2\)\)]], " = \[PlusMinus]2", StyleBox["a", FontSlant->"Italic"], ".\n\n(The + or the - on the right simply tells us which branch of the \ hyperbola we are on; it will disappear in a moment by squaring.) Now \ squaring, simplifying and substituting ", Cell[BoxData[ \(TraditionalForm\`b\^2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`c\^2 - a\^2\)]], ", we get\n\n", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)], FontSize->14, FontWeight->"Bold"], StyleBox["- ", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["1\n\n", FontSize->14, FontWeight->"Bold"], StyleBox["is the equation of a hyperbola with transverse axis the ", FontSize->12, FontWeight->"Bold"], StyleBox["x-axis", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", center the origin, foci at (-", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", 0) and (", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", 0) where ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`c\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`a\^2\)], FontWeight->"Bold"], StyleBox[" + ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`b\^2\)], FontWeight->"Bold"], StyleBox[". ", FontSize->12], StyleBox["The vertices are (", FontWeight->"Bold"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", 0) and (-", FontWeight->"Bold"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", 0).", FontWeight->"Bold"] }], "Text", FontSize->13], Cell[TextData[{ "Similarly \n\n", Cell[BoxData[ \(TraditionalForm\`y\^2\/a\^2\)], FontSize->14, FontWeight->"Bold"], StyleBox["- ", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`x\^2\/b\^2\)], FontSize->14, FontWeight->"Bold"], " = ", StyleBox["1\n\n", FontSize->14, FontWeight->"Bold"], StyleBox["is the equation of a hyperbola with transverse axis the ", FontSize->12, FontWeight->"Bold"], StyleBox["y-axis", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", center the origin, foci at (0, -", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") and (0, ", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") where ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`c\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`a\^2\)], FontWeight->"Bold"], StyleBox[" + ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`b\^2\)], FontWeight->"Bold"], StyleBox[". ", FontSize->12], StyleBox["The vertices are (0, ", FontWeight->"Bold"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") and (0, -", FontWeight->"Bold"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[").", FontWeight->"Bold"] }], "Text", FontSize->13], Cell[TextData[{ "Practice graphing these curves using the following programs. The first \ graph will be\n ", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)], FontWeight->"Bold"], StyleBox["- ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)], FontWeight->"Bold"], " = ", StyleBox["1", FontWeight->"Bold"], " and the second ", Cell[BoxData[ \(TraditionalForm\`y\^2\/a\^2\)], FontWeight->"Bold"], StyleBox["- ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`x\^2\/b\^2\)], FontWeight->"Bold"], " ", "= ", StyleBox["1", FontWeight->"Bold"], ". You choose ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], "." }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[a, b]\), ";", "\n", RowBox[{ StyleBox["a", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox[" ", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox["=", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox[" ", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox["10", FontColor->RGBColor[0, 0, 1]]}], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], RowBox[{ StyleBox["b", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox[" ", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox["=", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox[" ", FontColor->RGBColor[0, 0, 1], Background->RGBColor[1, 1, 0]], StyleBox["7", FontColor->RGBColor[0, 0, 1]]}], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(ImplicitPlot[ x\^2\/a\^2 - y\^2\/b\^2 - 1\ == \ 0, {x, \(-5\) - a, 5 + a}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]\), ";", "\n", \(ImplicitPlot[y\^2\/a\^2 - x\^2\/b\^2 - 1\ == \ 0, {x, \(-8\), 8}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]\), ";"}], "\n", "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t \ \t\t", StyleBox[\( (*shift - enter*) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["Now try the following exercises:", FontWeight->"Bold", FontVariations->{"Underline"->True}]], "Text", FontSize->13], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the center, transverse axis, foci, and vertices of each of the \ following hyperbolas. Then sketch the graph.\ni) ", Cell[BoxData[ \(TraditionalForm\`x\^2\/9\)]], " - ", Cell[BoxData[ \(TraditionalForm\`y\^2\/16\)]], " = 1\n\nii) ", Cell[BoxData[ \(TraditionalForm\`y\^2\/9\)]], " - ", Cell[BoxData[ \(TraditionalForm\`x\^2\/16\)]], " = 1\n\niii) 4", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " - ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = 4\n\niv) 4", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - 4", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " = 16" }], "Text", FontSize->13], Cell[BoxData[ \(Print[ "\"]; \n ImplicitPlot[x\^2\/9 - y\^2\/16 - 1\ == \ 0, {x, \(-8\), 8}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]; \n Print["\"]; \n ImplicitPlot[y\^2\/9 - x\^2\/16 - 1\ == \ 0, {x, \(-8\), 8}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]; \n Print[\*"\"\\""]; \n ImplicitPlot[x\^2 - y\^2\/4 - 1\ == \ 0, {x, \(-4\), 4}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]; \n Print[\*"\"\\""]; \n ImplicitPlot[y\^2\/4 - x\^2\/4 - 1\ == \ 0, {x, \(-4\), 4}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Asymptotes are useful graphing tools for hyperbolas. Recall that an \ asymptote of a curve is a line such that the graph gets arbitrarily near the \ line as x\[Rule]\[Infinity] or as x\[Rule]-\[Infinity].\n\n", StyleBox["The asymptotes of the hyperbola ", FontSize->12, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(x\^2\/a\^2\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["- ", FontSize->12, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(y\^2\/b\^2\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], StyleBox["1 are the lines y", FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], Cell[BoxData[ FormBox[ StyleBox[\(b\/a\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["x and y", FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], Cell[BoxData[ FormBox[ RowBox[{"-", StyleBox[\(b\/a\), FontSize->16]}], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["x.\nThe asymptotes of the hyperbola ", FontSize->12, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(y\^2\/a\^2\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["- ", FontSize->12, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(x\^2\/b\^2\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], StyleBox["1", FontSize->12, FontWeight->"Bold"], StyleBox[" ", FontSize->12], StyleBox["are the lines y", FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], Cell[BoxData[ FormBox[ StyleBox[\(a\/b\), FontSize->16], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["x and y", FontSize->12, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], Cell[BoxData[ FormBox[ RowBox[{"-", StyleBox[\(a\/b\), FontSize->16]}], TraditionalForm]], FontSize->12, FontWeight->"Bold"], StyleBox["x.", FontSize->12, FontWeight->"Bold"] }], "Text", FontSize->13], Cell[TextData[{ "For example, we draw below the graph of ", Cell[BoxData[ \(TraditionalForm\`x\^2\/16\)]], " - ", Cell[BoxData[ \(TraditionalForm\`y\^2\/4\)]], " = 1 with its asymptotes." }], "Text", FontSize->13], Cell[BoxData[ \(Clear[curv, asymp]; \n curv\ = \ ImplicitPlot[x\^2\/16 - y\^2\/9 - 1\ == \ 0, {x, \(-9\), 9}, DisplayFunction -> Identity, PlotStyle -> {Hue[0.9], Thickness[0.01]}]; \n asymp\ = \ Plot[{\(3\/4\) x, \(-\(3\/4\)\) x}, {x, \(-9\), 9}, DisplayFunction -> Identity, PlotStyle -> Hue[0.6]]; \n Show[{curv, asymp}, \ DisplayFunction -> $DisplayFunction]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ Remember the asymptotes are not actually a part of the graph, merely a tool \ to aid you in drawing a more accurate graph.\ \>", "Text", FontSize->13], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": For the hyperbola ", Cell[BoxData[ \(TraditionalForm\`y\^\(2\ \)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`\(\ x\^2\)\/4\)]], " = 1, determine the asymptotes and use them to help sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print["\"]; \n Clear[curv, asymp]; \n curv\ = \ ImplicitPlot[y\^2 - x\^2\/4 - 1\ == \ 0, {x, \(-6\), 6}, DisplayFunction -> Identity, PlotStyle -> {Hue[0.9], Thickness[0.01]}]; \n asymp\ = \ Plot[{\(1\/2\) x, \(-\(1\/2\)\) x}, {x, \(-7\), 7}, DisplayFunction -> Identity, PlotStyle -> Hue[0.6]]; \n Show[{curv, asymp}, \ DisplayFunction -> $DisplayFunction]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["2. Hyperbolas with center at (h,k)", "Subsection"], Cell[TextData[{ "This should come as no surprise by now. The equation of a hyperbola with \ transverse axis parallel to the ", StyleBox["x", FontSlant->"Italic"], " or ", StyleBox["y", FontSlant->"Italic"], " axis and center at (", StyleBox["h", FontSlant->"Italic"], ",", StyleBox["k", FontSlant->"Italic"], ") is given as follows:\n\n", StyleBox["The graph of", FontWeight->"Bold"], " ", Cell[BoxData[ \(TraditionalForm\`x\^2\/a\^2\)], FontWeight->"Bold"], StyleBox["- ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\/b\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontSize->12], StyleBox["1 has ", FontWeight->"Bold"], StyleBox["horizontal", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" transverse axis, center at (", FontWeight->"Bold"], StyleBox["h", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", ", FontWeight->"Bold"], StyleBox["k", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["), and foci at (", FontWeight->"Bold"], StyleBox["h", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" - ", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", ", FontWeight->"Bold"], StyleBox["k", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[") and (", FontWeight->"Bold"], StyleBox["h", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" +", FontWeight->"Bold"], StyleBox["c", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[", ", FontWeight->"Bold"], StyleBox["k", FontWeight->"Bold", FontSlant->"Italic"], StyleBox["), where ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`c\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`a\^2\)], FontWeight->"Bold"], StyleBox[" + ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`b\^2\)], FontWeight->"Bold"], StyleBox[". 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", "Sketch the graph, with its asymptotes." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \nClear[curv, asymp]; \n curv\ = \ ImplicitPlot[ \((y + 2)\)\^2\/4 - \((x - 3)\)\^2\/32 - 1\ == \ 0, {x, \(-10\), 15}, DisplayFunction -> Identity, PlotStyle -> {Hue[0.9], Thickness[0.01]}]; \n asymp\ = \ Plot[{\(\@2\/4\) \((x - 3)\) - 2, \(-\(\@2\/4\)\) \((x - 3)\) - 2}, {x, \(-10\), 15}, DisplayFunction -> Identity, PlotStyle -> Hue[0.6]]; \nShow[{curv, asymp}, \ DisplayFunction -> $DisplayFunction]; \)], "Input",\ FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the equation of the hyperbola with foci at (0, 4) and (0, -4) and \ an asymptote y = ", Cell[BoxData[ \(TraditionalForm\`\(1\/2\) x\)]], ". Sketch the graph, with its asymptotes. " }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n Clear[curv, asymp]; \n curv\ = \ ImplicitPlot[ \(5 y\^2\)\/16 - \(5 x\^2\)\/64 - 1\ == \ 0, {x, \(-8\), 8}, DisplayFunction -> Identity, PlotStyle -> {Hue[0.9], Thickness[0.01]}]; \n asymp\ = \ Plot[{1/2 x, \(-1\)/2 x}, {x, \(-8\), 8}, DisplayFunction -> Identity, PlotStyle -> Hue[0.6]]; \n Show[{curv, asymp}, \ DisplayFunction -> $DisplayFunction]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 4: Review exercises on conics ", StyleBox[" ", FontWeight->"Plain", FontSlant->"Italic"] }], "Section", FontSize->14, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["CLICK IN HERE FIRST!", FontVariations->{"Underline"->True}], " ", StyleBox["(Unless you've done it in the previous section.)", FontWeight->"Plain"] }], "Text", FontSize->13, FontWeight->"Bold"], Cell[BoxData[ \(Needs["\"]\)], "Input", FontColor->RGBColor[1, 0, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Ex1:", FontWeight->"Bold"], " Find the equation of the parabola with focus at (-3, 2) and directrix the \ line ", StyleBox["x", FontSlant->"Italic"], " = 1. Sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[\((y - 2)\) \((y - 2)\) + 8 \((x + 1)\)\ == \ 0, \ {x, \ \(-10\), \ 2}, \ AxesOrigin -> {0, 0}, PlotRange -> {{\(-10\), 2}, {\(-6\), \ 10}}, PlotStyle -> {Hue[0.9], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Ex2: ", FontWeight->"Bold"], "Find the equation of the ellipse with center at (1, 2), one focus at (1, \ 4), and containing the point (2, 2)", ". ", "Sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\
\""]; \n ImplicitPlot[ \((x - 1)\)\^2\/1 + \((y - 2)\)\^2\/5 - 1\ == \ 0, {x, \(-2\), 2}, PlotStyle -> {Hue[0.6], \ Thickness[0.011]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Ex3: ", FontWeight->"Bold"], "Find the equation of the hyperbola with center at (1, 4), one focus at \ (-2, 4), and one vertex at (0, 4). Sketch the graph." }], "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\<(h, k) = (1, 4), a = 1, c = 3, \!\(b\^2\) = 9 - 1 = 8. \!\(\((x \ - 1)\)\^2\/1\) - \!\(\((y - 4)\)\^2\/8\) = 1 is the equation.\>\""]; \n ImplicitPlot[ \((x - 1)\)\^2\/1 - \((y - 4)\)\^2\/8 - 1\ == \ 0, {x, \(-2\), 4}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Ex4: ", FontWeight->"Bold"], "Identify the conics given by the following equations. Give the foci, \ vertices, and directrix or asymptotes. Sketch the graphs.\n\t \t(a) ", Cell[BoxData[ FormBox[ StyleBox[\(y\^2\/25\), FontSize->16], TraditionalForm]]], "-", Cell[BoxData[ \(TraditionalForm\`\(\ x\^\(2\ \)\)\)]], " = 1\t\t\t(b) 9", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " + 4", Cell[BoxData[ \(TraditionalForm\`\(y\^2\ \)\)]], " = 36\n\t \t\t\n\t \t(c) 2", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - 4", StyleBox["y", FontSlant->"Italic"], " = ", StyleBox["x", FontSlant->"Italic"], " - 2\t\t\t(d) 4", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`9 y\^2\)]], " -16", StyleBox["x", FontSlant->"Italic"], " + 18", StyleBox["y", FontSlant->"Italic"], " = 11\n\t \t\t\n\t \t(e) 4", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " + 3", StyleBox["x", FontSlant->"Italic"], " - 16", StyleBox["y", FontSlant->"Italic"], " + 19 = 0\t\t(f) ", Cell[BoxData[ \(TraditionalForm\`\(\ x\^2\)\)]], " - ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - 2", StyleBox["x", FontSlant->"Italic"], " - 2", StyleBox["y", FontSlant->"Italic"], " = 1" }], "Text", FontSize->13], Cell[TextData[StyleBox["(a)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[{ \(Print[\*"\"\\""]\), RowBox[{ RowBox[{"ImplicitPlot", "[", " ", RowBox[{ RowBox[{ RowBox[{ StyleBox[ FormBox[\(y^2/25\), "TraditionalForm"], FontSize->11], "-", FormBox[\(\ x\^2\), "TraditionalForm"], "-", "1"}], " ", "==", " ", "0"}], ",", \({x, \(-2\), 2}\), ",", \(AspectRatio -> 1\), ",", \(PlotStyle -> {Hue[0.3], Thickness[0.01]}\)}], "]"}], ";"}]}], "Input", FontSize->11, FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["(b)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\<\!\(x\^2\/4\) + \!\(y\^2\/9\) = 1. Ellipse. Center (0, 0). \ Vertices (0, \[PlusMinus]3). Foci (0, \[PlusMinus]\!\(\@5\)).\>\""]; \n ImplicitPlot[\ x\^2\/4 + y\^2\/9 - 1\ == \ 0, {x, \(-2\), 2}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["(c)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[\ \((y - 1)\) \((y - 1)\) - \(1\/2\) x\ == \ 0, {x, \(-2\), 2}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["(d)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[\ \((x - 2)\)\^2\/9 + \((y + 1)\)\^2\/4 - 1\ == \ 0, {x, \(-3\), 7}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["(e)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\\""]; \n ImplicitPlot[\ \((y - 2)\)\^2 + \(3\/4\) \((x + 1)\)\ == \ 0, {x, \(-4\), 7}, AxesOrigin -> {0, 0}, PlotRange -> {{\(-4\), 1}, {\(-1\), 4}}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[StyleBox["(f)", FontWeight->"Bold"]], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\<(\!\(x\^2\) - 2x + 1) - (\!\(y\^2\) + 2y + 1) = 1 + 1 - 1; (x - \ 1\!\(\()\^2\)\) - (y + 1\!\(\()\^2\)\) = 1. Hyperbola. Center (1, -1). \ Vertices (0, -1) and (2, -1). Foci (1 \[PlusMinus] \!\(\@2\), -1). \ Asymptotes: y + 1 = \[PlusMinus](x - 1).\>\""]; \n ImplicitPlot[\ \((x - 1)\)\^2 - \((y + 1)\)\^2 - 1\ == \ 0, {x, \(-10\), 10}, AxesOrigin -> {0, 0}, PlotRange -> {{\(-4\), 6}, {\(-6\), 5}}, PlotStyle -> {Hue[0.3], Thickness[0.01]}]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Closed]], Cell[CellGroupData[{ Cell["Section 5: Rotation of axes; general form of a conic", "Section", FontSize->14, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["CLICK IN HERE FIRST!", FontVariations->{"Underline"->True}], " ", StyleBox["(Unless you've done it in the previous section.)", FontWeight->"Plain"] }], "Text", FontSize->14, FontWeight->"Bold"], Cell[BoxData[ \(Needs["\"]\)], "Input", FontColor->RGBColor[1, 0, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "We have now seen two different approaches to defining conics: as slices of \ a cone (as in the introduction), and as a collection of points with certain \ geometric properties. There is also an algebraic approach, which we present \ in this section.\n\tA general second degree polynomial with variables ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " looks like \n\t\t\t\t\t\t", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " = 0", StyleBox[" ", FontSlant->"Italic"], "\n(where ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], " and ", StyleBox["c", FontSlant->"Italic"], " are not all zero). We shall see that the graph of such an equation is \ always a conic--possibly tilted from the nice positions chosen in earlier \ sections. 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000Io`030?oo0?Oo000Io`030?oo0?Oo000Io`030?oo0?Oo000Io`030?oo0?Oo0000\ \>"], ImageRangeCache->{{{0, 274.688}, {199.438, 0}} -> {-1.00928, -5.00003, 0.0401131, 0.0401131}}], Cell[TextData[{ "We will ignore the 'degenerate cases', corresponding to those which arise \ from slicing a cone with a plane passing through the vertex of the cone.\n\t \ How can we show that the graph of ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " ", "= ", "0", StyleBox[" ", FontSlant->"Italic"], " is always a conic, and how can we tell what type of conic it is?" }], "Text", FontSize->13], Cell[CellGroupData[{ Cell[TextData[{ "1. The case ", StyleBox["b", FontSlant->"Italic"], " ", "= ", "0" }], "Subsection"], Cell[TextData[{ "Let's first look at the case when ", StyleBox["b", FontSlant->"Italic"], " = 0 in the general second degree polynomial. That is, ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " = 0", StyleBox[" ", FontSlant->"Italic"], ".\nThis case is nothing new! If ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["c", FontSlant->"Italic"], " are both nonzero, complete the square in ", StyleBox["x", FontSlant->"Italic"], " and in ", StyleBox["y", FontSlant->"Italic"], " to change it to the form ", StyleBox["a", FontSlant->"Italic"], "(", Cell[BoxData[ \(TraditionalForm\`\(x - h)\)\^2\)]], " + ", StyleBox["c", FontSlant->"Italic"], "(y -", Cell[BoxData[ \(TraditionalForm\`\(\ k)\)\^2\)]], " = constant,\nand we recognize the curve as an ellipse or hyperbola with \ center (", StyleBox["h", FontSlant->"Italic"], ", ", StyleBox["k", FontSlant->"Italic"], ") and axis parallel to the ", StyleBox["x", FontSlant->"Italic"], " or ", StyleBox["y", FontSlant->"Italic"], " axis (see sections 3 and 4). In fact, if ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["c", FontSlant->"Italic"], " have the same sign, it is an ellipse; if opposite signs, a hyperbola.\n\t\ On the other hand, if one of ", StyleBox["a", FontSlant->"Italic"], " or ", StyleBox["c", FontSlant->"Italic"], " is zero, then we can complete the square in only one of the variables. \ (Try this with ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - 2 ", StyleBox["x", FontSlant->"Italic"], " + 4 = 0.) We then get a parabola with directrix parallel to the ", StyleBox["x", FontSlant->"Italic"], " or ", StyleBox["y", FontSlant->"Italic"], " axis, as in section 1.\n\tThis gives us a very simple test to identify \ the type of conic with very little work:\n\t\t\n", StyleBox["Theorem:", FontSize->13, FontWeight->"Bold"], StyleBox[" ", FontSize->13], StyleBox["Except for a few degenerate cases, the equation ", FontSize->13, FontWeight->"Bold"], StyleBox["a ", FontSize->11, FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSize->11, FontSlant->"Italic"], StyleBox[" + c", FontSize->11, FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSize->11, FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSize->11, FontSlant->"Italic"], StyleBox[" = 0", FontSize->11], StyleBox[" ", FontSize->11, FontSlant->"Italic"], StyleBox[", where either ", FontSize->13, FontWeight->"Bold"], StyleBox["a", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" or ", FontSize->13, FontWeight->"Bold"], StyleBox["c", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" is not zero, is\n\t\t(a) a parabola if ", FontSize->13, FontWeight->"Bold"], StyleBox["a c", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" ", FontSize->13, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], StyleBox["0\n\t\t(b) an ellipse (or circle) if ", FontSize->13, FontWeight->"Bold"], StyleBox["a c", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" > 0\n\t\t(c) a hyperbola if ", FontSize->13, FontWeight->"Bold"], StyleBox["a c", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" < 0.", FontSize->13, FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "Of course, if you want more detailed information about the curve (like its \ center or axis), you must complete the square.\n\n\tUse the following program \ to test yourself on identifying these conics. You may choose the values of ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["c", FontSlant->"Italic"], ", ", StyleBox["d", FontSlant->"Italic"], ", ", StyleBox["e", FontSlant->"Italic"], ", and ", StyleBox["f", FontSlant->"Italic"], " (although only the values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["c", FontSlant->"Italic"], " matter).\n" }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[a, c, d, e, f, x, y, fct]\), ";", "\n", StyleBox[\(a\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(c\ = \ \(-1\)\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(d\ = \ \(-8\)\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(e\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(f\ = \ 1\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", RowBox[{"fct", " ", "=", " ", RowBox[{ RowBox[{ StyleBox["a", FontSize->14, FontWeight->"Bold"], FormBox[\(x\^2\), "TraditionalForm"]}], StyleBox["+", FontSize->14, FontWeight->"Bold"], RowBox[{ StyleBox["c", FontSize->14, FontWeight->"Bold"], FormBox[\(y\^2\), "TraditionalForm"]}], StyleBox["+", FontSize->14, FontWeight->"Bold"], StyleBox[\(d\ x\), FontSize->14, FontWeight->"Bold"], StyleBox["+", FontSize->14, FontWeight->"Bold"], StyleBox[\(e\ y\), FontSize->14, FontWeight->"Bold"], StyleBox["+", FontSize->14, FontWeight->"Bold"], StyleBox["f", FontSize->14, FontWeight->"Bold"]}]}], StyleBox[";", FontSize->14, FontWeight->"Bold"], "\n", \(If[a\ c\ == \ 0, Print["\", fct, \ "\< is a parabola, and here it is.\>"], If[a\ c > 0, Print["\", fct, \ "\< is an ellipse, and here it is.\>"], Print["\", fct, \ "\< is a hyperbola, and here it is.\>"]]]\), ";", "\n", \(ImplicitPlot[fct\ == \ 0, {x, \(-10\), 10}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]\), ";"}], "\n", " ", StyleBox[\( (*shift - enter*) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " Identify each of the following without completing the square. Then use \ the above program to check your answers.\n\n(a) ", Cell[BoxData[ RowBox[{ RowBox[{"2", StyleBox[ FormBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], "TraditionalForm"], FontWeight->"Plain", FontSlant->"Italic", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}]}], StyleBox["-", FontSize->14], RowBox[{ StyleBox["3", FontSize->14], FormBox[ SuperscriptBox[ StyleBox["y", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], "TraditionalForm"]}], StyleBox["+", FontSize->14], StyleBox[\(12\ x\), FontSize->14], StyleBox["-", FontSize->14], StyleBox["4", FontSize->14]}]]], "\n\n(b) ", Cell[BoxData[ RowBox[{ RowBox[{"3", StyleBox[ FormBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], "TraditionalForm"], FontFamily->"Times"]}], "-", RowBox[{"2", StyleBox[ FormBox[ SuperscriptBox[ StyleBox["y", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], "TraditionalForm"], FontFamily->"Times"]}], StyleBox["+", FontFamily->"Times"], StyleBox[\(5 x\), FontFamily->"Times"], StyleBox["-", FontFamily->"Times", FontSize->14], StyleBox[\(4 y\), FontSize->14]}]]], "\n\n(c) ", Cell[BoxData[ RowBox[{ FormBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], "TraditionalForm"], "-", \(5 x\), "+", \(4 y\), "+", "2"}]], FontFamily->"Times"] }], "Text", FontSize->13] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "2", ". ", "The case ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0:" }], "Subsection", FontSize->13], Cell[TextData[{ "Understanding the case when ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0 requires a rotation of the standard ", StyleBox["xy", FontSlant->"Italic"], " coordinate plane.\nIf we rotate the ", StyleBox["xy", FontSlant->"Italic"], "-plane counterclockwise through an angle \[Theta], fixing the origin, we \ take the ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " axis to new axes, which we will call ", StyleBox["x'", FontSlant->"Italic"], " and ", StyleBox["y'", FontSlant->"Italic"], " as shown below." }], "Text", FontSize->13], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .61803 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations 0.34127 0.15873 0.0147151 0.147151 [ [1.025 .01472 0 -6.28125 ] [1.025 .01472 10 6.28125 ] [.34127 .64303 -5 0 ] [.34127 .64303 5 12.5625 ] [ 0 0 0 0 ] [ 1 .61803 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 g .25 Mabswid 0 .01472 m 1 .01472 L s gsave 1.025 .01472 -61 -10.2813 Mabsadd m 1 1 Mabs scale currentpoint translate 0 20.5625 translate 1 -1 scale gsave 0.000000 0.000000 0.000000 setrgbcolor 1.000000 setlinewidth gsave newpath 61.000000 16.562500 moveto 460.000000 16.562500 lineto 460.000000 4.000000 lineto 61.000000 4.000000 lineto 61.000000 16.562500 lineto closepath clip newpath 63.000000 12.812500 moveto %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10.000000 scalefont [1 0 0 -1 0 0 ] makefont setfont 0.000000 0.000000 0.000000 setrgbcolor (x) show 1.000000 setlinewidth grestore grestore %%DocumentNeededResources: font Courier %%DocumentSuppliedResources: %%DocumentNeededFonts: Courier %%DocumentSuppliedFonts: %%DocumentFonts: font Courier grestore .34127 0 m .34127 .61803 L s gsave .34127 .64303 -66 -4 Mabsadd m 1 1 Mabs scale currentpoint translate 0 20.5625 translate 1 -1 scale gsave 0.000000 0.000000 0.000000 setrgbcolor 1.000000 setlinewidth gsave newpath 61.000000 16.562500 moveto 460.000000 16.562500 lineto 460.000000 4.000000 lineto 61.000000 4.000000 lineto 61.000000 16.562500 lineto closepath clip newpath 63.000000 12.812500 moveto %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10.000000 scalefont [1 0 0 -1 0 0 ] makefont setfont 0.000000 0.000000 0.000000 setrgbcolor (y) show 1.000000 setlinewidth grestore grestore %%DocumentNeededResources: font Courier %%DocumentSuppliedResources: %%DocumentNeededFonts: Courier %%DocumentSuppliedFonts: %%DocumentFonts: font Courier grestore 0 0 m 1 0 L 1 .61803 L 0 .61803 L closepath clip newpath .2 0 1 r .5 Mabswid .34127 .01472 m .97619 .30902 L s .34127 .01472 m .02381 .60332 L s 0 g .03 w .81746 .30902 Mdot .5 Mabswid newpath matrix currentmatrix 0.15873 0.147151 scale 2.15 .10003 1 0 26.5623 arc setmatrix s % End of Graphics MathPictureEnd \ \>"], "Graphics", ImageSize->{288, 177.938}, ImageMargins->{{43, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, FontSize->11, ImageCache->GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHgOol00aPOOomoo`0>Ool00`00Oomoo`0oOol3 61mlOol004ioo`0301moogoo00ioo`03001oogoo049oo`8H7gYoo`00CGoo00goo00Woo007oo00Goo00<007ooOol0 _Woo000ROol00aPOOomoo`0jOol00`00Oomoo`2nOol0029oo`0361moogoo03Yoo`03001oogoo0;io o`008Goo00goo00<007ooOol0_Woo000QOol00aPOOomoo`0kOol00`00Oomoo`2nOol0 021oo`0361moogoo03aoo`03001oogoo0;ioo`0087oo00"], ImageRangeCache->{{{0, 287}, {176.938, 0}} -> {-2.38905, -0.169635, 0.0250383, 0.0270085}}], Cell[BoxData[ \("The rotated axes are in blue. The arc represents the angle of \ rotation."\)], "Print", FontSize->11], Cell[TextData[{ "\nNotice that we can describe the location of the point ", StyleBox["P", FontSlant->"Italic"], " using ", StyleBox["x'", FontSlant->"Italic"], " and ", StyleBox["y'", FontSlant->"Italic"], ", as well as ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], ". In other words, we can locate ", StyleBox["P", FontSlant->"Italic"], " by travelling so many units \nalong the ", StyleBox["x'", FontSlant->"Italic"], " axis and then travelling so many units parallel to the ", StyleBox["y'", FontSlant->"Italic"], " axis. So for some numbers ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], ", ", StyleBox["P", FontSlant->"Italic"], " = (", StyleBox["x,y", FontSlant->"Italic"], ") in the original coordinate plane, and \nfor some numbers ", StyleBox["x'", FontSlant->"Italic"], " and ", StyleBox["y'", FontSlant->"Italic"], ", ", StyleBox["P", FontSlant->"Italic"], " = (", StyleBox["x',y'", FontSlant->"Italic"], ") in the new coordinate plane. What is the relationship between the ", StyleBox["x,y", FontSlant->"Italic"], " coordinates and the ", StyleBox["x',y'", FontSlant->"Italic"], " coordinates of the \npoint ", StyleBox["P", FontSlant->"Italic"], "? Some simple right triangle trigonometry gives the answer.\n\nDraw two \ right triangles: one by dropping a perpendicular from ", StyleBox["P", FontSlant->"Italic"], " to the ", StyleBox["x", FontSlant->"Italic"], " axis and another by dropping a perpendicular to the ", StyleBox["x'", FontSlant->"Italic"], " axis.\n\n\t\n" }], "Text", FontSize->13], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .61803 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations 0.34127 0.15873 0.0147151 0.147151 [ [1.025 .01472 0 -6.28125 ] [1.025 .01472 10 6.28125 ] [.34127 .64303 -5 0 ] [.34127 .64303 5 12.5625 ] [ 0 0 0 0 ] [ 1 .61803 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 g .25 Mabswid 0 .01472 m 1 .01472 L s gsave 1.025 .01472 -61 -10.2813 Mabsadd m 1 1 Mabs scale currentpoint translate 0 20.5625 translate 1 -1 scale gsave 0.000000 0.000000 0.000000 setrgbcolor 1.000000 setlinewidth gsave newpath 61.000000 16.562500 moveto 460.000000 16.562500 lineto 460.000000 4.000000 lineto 61.000000 4.000000 lineto 61.000000 16.562500 lineto closepath clip newpath 63.000000 12.812500 moveto %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10.000000 scalefont [1 0 0 -1 0 0 ] makefont setfont 0.000000 0.000000 0.000000 setrgbcolor (x) show 1.000000 setlinewidth grestore grestore %%DocumentNeededResources: font Courier %%DocumentSuppliedResources: %%DocumentNeededFonts: Courier %%DocumentSuppliedFonts: %%DocumentFonts: font Courier grestore .34127 0 m .34127 .61803 L s gsave .34127 .64303 -66 -4 Mabsadd m 1 1 Mabs scale currentpoint translate 0 20.5625 translate 1 -1 scale gsave 0.000000 0.000000 0.000000 setrgbcolor 1.000000 setlinewidth gsave newpath 61.000000 16.562500 moveto 460.000000 16.562500 lineto 460.000000 4.000000 lineto 61.000000 4.000000 lineto 61.000000 16.562500 lineto closepath clip newpath 63.000000 12.812500 moveto %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10.000000 scalefont [1 0 0 -1 0 0 ] makefont setfont 0.000000 0.000000 0.000000 setrgbcolor (y) show 1.000000 setlinewidth grestore grestore %%DocumentNeededResources: font Courier %%DocumentSuppliedResources: %%DocumentNeededFonts: Courier %%DocumentSuppliedFonts: %%DocumentFonts: font Courier grestore 0 0 m 1 0 L 1 .61803 L 0 .61803 L closepath clip newpath .2 0 1 r .5 Mabswid .34127 .01472 m .97619 .30902 L s .34127 .01472 m .02381 .60332 L s 0 g .03 w .81746 .30902 Mdot .5 Mabswid .81746 .30902 m .81746 .01472 L s .2 0 1 r .81746 .30902 m .84921 .25016 L s 1 0 0 r .34127 .01472 m .81746 .30902 L s % End of Graphics MathPictureEnd \ \>"], "Graphics", ImageSize->{288, 177.938}, ImageMargins->{{43, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, FontSize->11, ImageCache->GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHgOol005ioo`0461l007ooOol2 O0000aPOOomoo`1`Ool00`00Oomoo`16Ool005eoo`0361moo`0000Aoo`03O00H7aPO071oo`03001o ogoo04Ioo`00GGoo007oo00<007ooOol0AWoo001?Ool00aPOOomoo`0=Ool00`00Oomoo`0]Ool2O00>Ool2 61lfOol00`00Oomoo`16Ool004ioo`0361moogoo00ioo`03001oogoo02moo`9l00ioo`Woo0W`04goo0QPO97oo00<007ooOol0AWoo001:Ool00aPOOomoo`0BOol00`00 Oomoo`0lOol2O00COol261lROol00`00Oomoo`16Ool004Uoo`0361moogoo01=oo`03001oogoo03io o`03O01oogoo019oo`0361l07goo01moo`03001oogoo04Ioo`00BGoo00Ool00`00Oomo o`16Ool004Eoo`0361moogoo01Moo`03001oogoo04aoo`9l01Qoo`Goo007oo00<007ooOol0_Woo000SOol00aPOOomoo`0i Ool00`00Oomoo`2nOol0029oo`0361moogoo03Yoo`03001oogoo0;ioo`008Woo00Woo 00<007ooOol0_Woo000QOol00aPOOomoo`0kOol00`00Oomoo`2nOol0025oo`0361moogoo03]oo`03 001oogoo0;ioo`0087oo00"], ImageRangeCache->{{{0, 287}, {176.938, 0}} -> {-2.38905, -0.169635, 0.0250383, 0.0270085}}], Cell[TextData[{ "\n\t\n\t Both right triangles have the same hypoteneuse--the ", StyleBox["(red)", FontColor->RGBColor[1, 0, 0]], " line segment joining ", StyleBox["P", FontSlant->"Italic"], " to the origin--with length ", StyleBox["r", FontSlant->"Italic"], ". The legs of the right triangles have lengths ", StyleBox["x, y, x', y',", FontSlant->"Italic"], " respectively. We let \[Alpha] be the angle between the positive ", StyleBox["x' ", FontSlant->"Italic"], "axis and the hypoteneuse. Then by the right triangle definitions of sine \ and cosine we have\n\t", StyleBox["x'", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " cos(\[Alpha])\t\t", StyleBox["y'", FontSlant->"Italic"], " = ", StyleBox["r ", FontSlant->"Italic"], "sin(\[Alpha])\n\t", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " cos(\[Theta] + \[Alpha])\t", StyleBox["y", FontSlant->"Italic"], " = ", StyleBox["r", FontSlant->"Italic"], " sin(\[Theta] + \[Alpha])\nUsing the sum formula, we get\n", StyleBox["\n\t", FontSize->14], StyleBox["x", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], " = ", StyleBox["x'", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" cos(\[Theta]) - ", FontSize->14, FontWeight->"Bold"], StyleBox["y'", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" sin(\[Theta])\t\t", FontSize->14, FontWeight->"Bold"], StyleBox["y", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], " = ", StyleBox["x'", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" sin(\[Theta]) + ", FontSize->14, FontWeight->"Bold"], StyleBox["y'", FontSize->14, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" cos(\[Theta])", FontSize->14, FontWeight->"Bold"], StyleBox["\n\t", FontWeight->"Bold"], "\nthe relationship we were seeking.\n\tBefore returning to the question \ about conics, let's practice finding new coordinates of a point under \ rotation by \[Theta]. Given the original coordinates ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " of a point ", StyleBox["P", FontSlant->"Italic"], " and the angle \[Theta], your task is to find the new coordinates ", StyleBox["x'", FontSlant->"Italic"], " and ", StyleBox["y'", FontSlant->"Italic"], " of ", StyleBox["P", FontSlant->"Italic"], ". Try it on your own, choosing ", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], " and \[Theta], and then use the following program to check your answers.\t\ " }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[x, y, X, Y, \[Theta], sol]\), ";", "\n", StyleBox[\(x\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(y\ = \ 5\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(\[Theta]\ = \ \[Pi]\/3\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(sol\ = \ Solve[{x\ == \ X\ Cos[\[Theta]] - Y\ Sin[\[Theta]], y\ == \ X\ Sin[\[Theta]] + Y\ Cos[\[Theta]]}, {X, Y}]\), ";", "\n", \(Print[ "\", {sol[\([1, 1, 2]\)], sol[\([1, 2, 2]\)]}]\)}], "\n", "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox[\( (*shift - enter*) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontSize->11, FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " Find the new ", StyleBox["x'", FontSlant->"Italic"], ", ", StyleBox["y'", FontSlant->"Italic"], " coordinates of the point ", StyleBox["P", FontSlant->"Italic"], " = (1, 3) when the axes are rotated by the given angle \[Theta]. Then \ use the above program to check your answers.\n\n(a) \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], "\n\n(b) \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], "\n\n(c) \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\(5 \[Pi]\)\/3\)]], "\n\n(d) \[Theta] = -", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]] }], "Text", FontSize->13], Cell[TextData[{ "\tNow to return to conics--by wisely choosing the angle \[Theta] through \ which to rotate, we can transform a", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " + bxy + c", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " + dx + ey + f = 0, where ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0, into an equation in ", StyleBox["x' ", FontSlant->"Italic"], "and ", StyleBox["y'", FontSlant->"Italic"], " ", StyleBox["without an x'y' term", FontSlant->"Italic"], ". The angle that works is given by cot(2\[Theta]) = ", Cell[BoxData[ \(TraditionalForm\`\(a\ - \ c\)\/b\)], FontSize->16], ". To summarize:\n\n", StyleBox["\t", FontSize->14, FontWeight->"Bold"], StyleBox["To eliminate the ", FontWeight->"Bold"], StyleBox["xy", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" term from ", FontSize->13, FontWeight->"Bold"], StyleBox["a ", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" + b x y + c", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" = ", FontSize->11], StyleBox["0", FontSize->11, FontWeight->"Bold"], StyleBox[", substitute\n\t", FontSize->13, FontWeight->"Bold"], StyleBox["\n\t", FontSize->13], StyleBox["x", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" = ", FontSize->11], StyleBox["x'", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" cos(\[Theta]) - ", FontSize->13, FontWeight->"Bold"], StyleBox["y'", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" sin(\[Theta])\t\t", FontSize->13, FontWeight->"Bold"], StyleBox["y", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" = ", FontSize->11], StyleBox["x'", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" sin(\[Theta]) + ", FontSize->13, FontWeight->"Bold"], StyleBox["y'", FontSize->13, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" cos(\[Theta])\n\t", FontSize->13, FontWeight->"Bold"], StyleBox["\n\t", FontSize->13], StyleBox["where \[Theta] is such that cot(2\[Theta])", FontSize->13, FontWeight->"Bold"], StyleBox[" = ", FontSize->11], Cell[BoxData[ \(TraditionalForm\`\(a\ - \ c\)\/b\)], FontSize->17, FontWeight->"Bold"], StyleBox[".", FontSize->17] }], "Text"], Cell[TextData[ "You will need to be able to find the angle \[Theta], so practice!"], "Text", FontSize->13], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " Find the angle \[Theta] that will eliminate the ", StyleBox["x y", FontSlant->"Italic"], " term in each of the following equations. Check your answers with the \ above program.\n\n(a) ", StyleBox["x y", FontSlant->"Italic"], " = 4 \n\n(b) 3 ", Cell[BoxData[ \(TraditionalForm\`x\^\(2\ \)\)]], "- 4 ", StyleBox["x y", FontSlant->"Italic"], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " + 4 = 0\n\n(c) 6", Cell[BoxData[ \(TraditionalForm\`\(\ x\^2\ \)\)]], "- 4", Cell[BoxData[ \(TraditionalForm\`\@3\)]], " ", StyleBox["x y", FontSlant->"Italic"], " + 2 ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " - ", StyleBox["x", FontSlant->"Italic"], " - ", Cell[BoxData[ \(TraditionalForm\`\@3\)]], " ", StyleBox["y", FontSlant->"Italic"], " = 0" }], "Text", FontSize->13], Cell["\<\ Solutions: (a) \ \>", "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]\)], "Input", FontSize->11, FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell["(b)", "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]\)], "Input", FontSize->11, FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell["(c)", "Text", FontSize->13], Cell[BoxData[ \(Print[ \*"\"\\""]\)], "Input", FontSize->11, FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Why is eliminating this term important? What claim did we make earlier \ that we have proven? Think about this for a moment before reading on.\n\t \ Recall we began this section by claiming that the graph of any second degree \ polynomial in variables ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], ", ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f ", FontSlant->"Italic"], "= 0, is a conic curve (except for a few exceptional cases). We have now \ found a change of variables that transforms such an equation into one without \ an ", StyleBox["x'y'", FontSlant->"Italic"], " term. But by the first part of this section, we know that the graph of \ this transformed equation is a conic--we just complete the square to put it \ in standard form. So the graph of ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f ", FontSlant->"Italic"], "= 0 is a conic, merely tilted if ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual]0.\n\tThis means that if you are given an equation ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " = 0, you could first find the angle \[Theta] that would eliminate the ", StyleBox["x y", FontSlant->"Italic"], " term (if there is one), make the appropriate substitutions and then \ complete the square to put it in standard form for a parabola, ellipse, or \ hyperbola and graph it. Warning: the computations involved in making the \ substitutions is generally long and tedious! However we'll use ", StyleBox["Mathematica", FontSlant->"Italic"], " to help us out.\n\tOur original equation looks like ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " = 0; you may choose values for ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["c", FontSlant->"Italic"], ", ", StyleBox["d", FontSlant->"Italic"], ", ", StyleBox["e", FontSlant->"Italic"], ", ", StyleBox["f", FontSlant->"Italic"], " with ", StyleBox["b", FontSlant->"Italic"], " \[NotEqual] 0. ", StyleBox["You", FontVariations->{"Underline"->True}], " will need to find the appropriate value of \[Theta] and plug it in. Let \ ", StyleBox["Mathematica", FontSlant->"Italic"], " do the rest! The program will make the substitutions called for." }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ \(Clear[a, b, c, d, e, f, \[Theta], x, y, fct, newfct]\), ";", "\n", StyleBox[\(a\ = \ 4\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(b\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(c\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(d\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(e\ = \ \(-1\)\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(f\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[ RowBox[{"\[Theta]", " ", "=", " ", FractionBox["\[Pi]", StyleBox["6", FontColor->GrayLevel[0]]]}], FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", \(fct\ = \ a\ x\^2 + b\ x\ y\ + c\ y\^2 + d\ x\ + \ e\ y\ + \ f\), ";", "\n", \(newfct\ = \ fct /. {x -> x'\ Cos[\[Theta]] - y'\ Sin[\[Theta]], y -> x'\ Sin[\[Theta]] + y'\ Cos[\[Theta]]}\), ";", "\n", \(Simplify[newfct]\)}], "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t \ "}]], "Input", FontSize->11, FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tFortunately, just as in the ", StyleBox["b", FontSlant->"Italic"], " ", "= ", "0 case, we can ", StyleBox["identify", FontSlant->"Italic"], " the ", StyleBox["type", FontSlant->"Italic"], " of conic without going through this mess." }], "Text", FontSize->13], Cell[TextData[{ StyleBox[ "Theorem: Except for a few degenerate cases, the graph of \n\t\t\t", FontSize->13, FontWeight->"Bold"], StyleBox["a ", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" + b x y + c", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSize->11, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" = ", FontSize->11], StyleBox["0", FontSize->11, FontWeight->"Bold"], StyleBox[" \nis\n(a) a parabola if ", FontSize->13, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(b\^2 - 4 a\ c\), FontWeight->"Bold"], TraditionalForm]], FontSize->13], StyleBox[" = ", FontSize->11], StyleBox["0\n(b) an ellipse (or circle) if ", FontSize->13, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(b\^2 - 4 a\ c\ \), FontWeight->"Bold"], TraditionalForm]], FontSize->13], StyleBox["< 0\n(c) a hyperbola if ", FontSize->13, FontWeight->"Bold"], Cell[BoxData[ FormBox[ StyleBox[\(b\^2 - 4 a\ c\), FontWeight->"Bold"], TraditionalForm]], FontSize->13], StyleBox[" > 0.", FontSize->13, FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "Pracatice identifying conics with the following program. Again you may \ choose values for ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["c", FontSlant->"Italic"], ", ", StyleBox["d", FontSlant->"Italic"], ", ", StyleBox["e", FontSlant->"Italic"], ", ", StyleBox["f", FontSlant->"Italic"], " in ", StyleBox["a ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontSlant->"Italic"], StyleBox[" + b x y + c", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontSlant->"Italic"], StyleBox[" + d x + e y + f", FontSlant->"Italic"], " = 0 . Your goal is to identify the type of curve.\n(Question: can you \ let ", StyleBox["b", FontSlant->"Italic"], " = 0 here? Why or why not?)" }], "Text", FontSize->13], Cell[BoxData[ RowBox[{ RowBox[{\(Clear[a, b, c, d, e, f, x, y, fct]\), ";", "\n", StyleBox[\(a\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(b\ = \ 0\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(c\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(d\ = \ 2\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(e\ = \ 4\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], StyleBox["\n", FontColor->GrayLevel[0]], StyleBox[\(f\ = \ 1\), FontColor->GrayLevel[0]], StyleBox[";", FontColor->GrayLevel[0]], "\n", RowBox[{"fct", "=", " ", RowBox[{ RowBox[{"a", FormBox[\(x\^2\), "TraditionalForm"]}], "+", \(b\ x\ y\), "+", RowBox[{"c", FormBox[\(y\^2\), "TraditionalForm"]}], "+", \(d\ x\), "+", \(e\ y\), "+", "f"}]}], ";", "\n", \(If[b\^2 - 4\ a\ c\ == \ 0, Print["\", fct, \ "\< is a parabola.\>"], If[b\^2 - 4\ a\ c > 0, Print["\", fct, \ "\"], Print["\", fct, \ "\"]]]\), ";", "\n", \(ImplicitPlot[fct\ == \ 0, {x, \(-10\), 10}, PlotStyle -> {Hue[0.9], Thickness[0.008]}]\), ";"}], "\n", "\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t", StyleBox[\( (*\ shift - enter\ *) \), FontColor->RGBColor[0, 0, 1]]}]], "Input", FontSize->11, FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise: ", FontWeight->"Bold"], "Identify the type of conic represented by each of the following equations. \ Then use the above program to check your answers. 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