# 1.5 Conic sections  (Page 9/23)

 Page 9 / 23
1. $4AC-{B}^{2}>0.$ If so, the graph is an ellipse.
2. $4AC-{B}^{2}=0.$ If so, the graph is a parabola.
3. $4AC-{B}^{2}<0.$ If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is $xy=1.$ This equation can be solved for y to obtain $y=\frac{1}{x}.$ The graph of this function is called a rectangular hyperbola as shown.

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle $\theta$ of rotation of the conic section, we use the formula $\text{cot}\phantom{\rule{0.2em}{0ex}}2\theta =\frac{A-C}{B}.$ In this case $A=C=0$ and $B=1,$ so $\text{cot}\phantom{\rule{0.2em}{0ex}}2\theta =\left(0-0\right)\text{/}1=0$ and $\theta =45\text{°}.$ The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled ${A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{F}^{\prime },$ and are given by the formulas

$\begin{array}{ccc}\hfill {A}^{\prime }& =\hfill & A\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\theta +B\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta +C\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\theta \hfill \\ \hfill {B}^{\prime }& =\hfill & 0\hfill \\ \hfill {C}^{\prime }& =\hfill & A\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\theta -B\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +C\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\theta \hfill \\ \hfill {D}^{\prime }& =\hfill & D\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +E\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ \hfill {E}^{\prime }& =\hfill & \text{−}D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta +E\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ \hfill {F}^{\prime }& =\hfill & F.\hfill \end{array}$

The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant $4AC-{B}^{2}.$
2. Determine $\theta$ using the formula $\text{cot}\phantom{\rule{0.2em}{0ex}}2\theta =\frac{A-C}{B}.$
3. Calculate ${A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{F}^{\prime }.$
4. Rewrite the original equation using ${A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{F}^{\prime }.$
5. Draw a graph using the rotated equation.

## Identifying a rotated conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-256=0.$

In this equation, $A=13,B=-6\sqrt{3},C=7,D=0,E=0,$ and $F=-256.$ The discriminant of this equation is $4AC-{B}^{2}=4\left(13\right)\left(7\right)-{\left(-6\sqrt{3}\right)}^{2}=364-108=256.$ Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use $\text{cot}\phantom{\rule{0.2em}{0ex}}2\theta =\frac{A-C}{B}.$ This gives

$\begin{array}{cc}\hfill \text{cot}\phantom{\rule{0.2em}{0ex}}2\theta & =\frac{A-C}{B}\hfill \\ & =\frac{13-7}{-6\sqrt{3}}\hfill \\ & =-\frac{\sqrt{3}}{3}.\hfill \end{array}$

Therefore $2\theta ={120}^{\text{o}}$ and $\theta ={60}^{\text{o}},$ which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

$\begin{array}{ccc}\hfill {A}^{\prime }& =\hfill & A\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\theta +B\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta +C\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\theta \hfill \\ & =\hfill & 13{\text{cos}}^{2}60+\left(-6\sqrt{3}\right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}60\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}60+7{\text{sin}}^{2}60\hfill \\ & =\hfill & 13{\left(\frac{1}{2}\right)}^{2}-6\sqrt{3}\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+7{\left(\frac{\sqrt{3}}{2}\right)}^{2}\hfill \\ & =\hfill & 4,\hfill \\ \hfill {B}^{\prime }& =\hfill & 0,\hfill \\ \hfill {C}^{\prime }& =\hfill & A\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\theta -B\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +C\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}\theta \hfill \\ & =\hfill & 13{\text{sin}}^{2}60+\left(-6\sqrt{3}\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}60\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}60=7{\text{cos}}^{2}60\hfill \\ \hfill & =\hfill & {\left(\frac{\sqrt{3}}{2}\right)}^{2}+6\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)+7{\left(\frac{1}{2}\right)}^{2}\hfill \\ & =\hfill & 16,\hfill \\ \hfill {D}^{\prime }& =\hfill & D\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta +E\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & =\hfill & \left(0\right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}60+\left(0\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}60\hfill \\ & =\hfill & 0,\hfill \\ \hfill {E}^{\prime }& =\hfill & \text{−}D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta +E\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \hfill \\ & =\hfill & \text{−}\left(0\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}60+\left(0\right)\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}60\hfill \\ & =\hfill & 0,\hfill \\ \hfill {F}^{\prime }& =\hfill & F\hfill \\ & =\hfill & -256.\hfill \end{array}$

The equation of the conic in the rotated coordinate system becomes

$\begin{array}{}\\ \hfill 4{\left({x}^{\prime }\right)}^{2}+16{\left({y}^{\prime }\right)}^{2}& =\hfill & 256\hfill \\ \hfill \frac{{\left({x}^{\prime }\right)}^{2}}{64}+\frac{{\left({y}^{\prime }\right)}^{2}}{16}& =\hfill & 1.\hfill \end{array}$

A graph of this conic section appears as follows.

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$3{x}^{2}+5xy-2{y}^{2}-125=0.$

The conic is a hyperbola and the angle of rotation of the axes is $\theta =22.5\text{°}.$

## Key concepts

• The equation of a vertical parabola in standard form with given focus and directrix is $y=\frac{1}{4p}{\left(x-h\right)}^{2}+k$ where p is the distance from the vertex to the focus and $\left(h,k\right)$ are the coordinates of the vertex.
• The equation of a horizontal ellipse in standard form is $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ where the center has coordinates $\left(h,k\right),$ the major axis has length 2 a, the minor axis has length 2 b , and the coordinates of the foci are $\left(h±c,k\right),$ where ${c}^{2}={a}^{2}-{b}^{2}.$
• The equation of a horizontal hyperbola in standard form is $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ where the center has coordinates $\left(h,k\right),$ the vertices are located at $\left(h±a,k\right),$ and the coordinates of the foci are $\left(h±c,k\right),$ where ${c}^{2}={a}^{2}+{b}^{2}.$
• The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
• The polar equation of a conic section with eccentricity e is $r=\frac{ep}{1±e\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta }$ or $r=\frac{ep}{1±e\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta },$ where p represents the focal parameter.
• To identify a conic generated by the equation $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0,$ first calculate the discriminant $D=4AC-{B}^{2}.$ If $D>0$ then the conic is an ellipse, if $D=0$ then the conic is a parabola, and if $D<0$ then the conic is a hyperbola.

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