7.5 Conic sections  (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 $\text{cot}2\theta =\frac{A-C}{B}.$ In this case $A=C=0$ and $B=1,$ so $\text{cot}2\theta =(0-0)\text{/}1=0$ 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 $A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime },\text{and}{F}^{\prime },$ and are given by the formulas

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 $\text{cot}2\theta =\frac{A-C}{B}.$
3. Calculate $A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime },\text{and}{F}^{\prime }.$
4. Rewrite the original equation using $A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime },\text{and}{F}^{\prime }.$
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

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

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

$\begin{array}{cc}\hfill \text{cot}2\theta & =\frac{A-C}{B}\hfill \\ & =\frac{13-7}{\mathrm{-6}\sqrt{3}}\hfill \\ & =-\frac{\sqrt{3}}{3}.\hfill \end{array}$

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:

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

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.

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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 $\left(h\pm c,k\right),$ 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 $\left(h\pm a,k\right),$ and the coordinates of the foci are $\left(h\pm c,k\right),$ 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 $r=\frac{ep}{1\pm e\text{cos}\theta }$ or $r=\frac{ep}{1\pm e\text{sin}\theta },$ 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.