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Precalculus : Find the Polar Equation of a Conic Section

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Example Question #11 : Find The Polar Equation Of A Conic Section

Write the polar equation for $y=\frac{1}{4} x^2 - 3$

Possible Answers:

$r = \frac{ 2 \sqrt 3 }{ \cos \theta }$

$r = \frac{ 3 \sin \theta + 1 }{ \cos ^2 \theta }$

$r = \frac{\sin \theta - \sqrt{ \sin ^2 \theta - 3 \cos ^2 \theta }}{ 2 \cos ^2 \theta }$

$r = \frac{ \sin \theta + \frac{1}{3}}{\frac{1}{2} \cos ^2 \theta }$

$r = \frac{ \sin \theta + \sqrt{ \sin ^2 \theta + 3 \cos \theta }}{\frac{1}{2} \cos ^2 \theta}$

Correct answer:

$r = \frac{ \sin \theta + \sqrt{ \sin ^2 \theta + 3 \cos \theta }}{\frac{1}{2} \cos ^2 \theta}$

Explanation:

To convert, make the substitutions $x = r \cos \theta$ and $y = r \sin \theta$

$r \sin \theta = \frac{1}{4} r^2 \cos ^2 \theta - 3$ subtract $r \sin \theta$ from both sides

$0 = \frac{1}{4} r^2 \cos^2 \theta - r \sin \theta - 3$

Now we can solve using the quadratic formula:

$r = \frac{\sin \theta \pm \sqrt{\sin^2\theta - 4(\frac{1}{4}\cos ^2 \theta)(-3)}}{2(\frac{1}{4} \cos ^2 \theta)}$

$r = \frac{ \sin \theta \pm \sqrt{ \sin ^2 \theta + 3 \cos ^2 \theta }}{ \frac{1}{2} \cos ^2 \theta}$

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Example Question #41 : Conic Sections

Write the equation y = 3x^2 + 2 in polar form

Possible Answers:

$r = \frac{ 2\sin \theta - \sqrt{ 24} \cos \theta }{ 6 \cos ^2 \theta }$

$r = \frac{ \sin \theta + \sqrt{ \sin ^2 \theta - 24 \cos ^2 \theta }}{ 6 \cos ^2 \theta }$

$r = \frac{ \sin \theta + \sqrt{ \sin ^2 \theta+24 \cos ^2 \theta }}{ 6 \cos ^2 \theta }$

$r = \frac{ \sin \theta + 5 \cos \theta }{ 3 \cos ^2 \theta }$

$r = \frac{ \sin \theta - \sqrt{ \sin ^2 \theta + 24 \cos ^2 \theta }}{ 3 \cos ^2 \theta }$

Correct answer:

$r = \frac{ \sin \theta + \sqrt{ \sin ^2 \theta - 24 \cos ^2 \theta }}{ 6 \cos ^2 \theta }$

Explanation:

Make the substitutions $x = r \cos \theta$ and $y = r \sin \theta$ :

$r \sin \theta = 3 r^2 \cos ^2 \theta + 2$ subtract $r \sin \theta$ from both sides

$0=3 r^2 \cos ^2 \theta - r \sin \theta + 2$

We can solve for r using the quadratic formula:

$r = \frac{ \sin \theta \pm \sqrt{ \sin^ 2 \theta - 4(3 \cos ^2 \theta ) (2)}}{2( 3 \cos ^2 \theta )}$

$r = \frac{\sin \theta \pm \sqrt{ \sin ^2 \theta - 24 \cos ^2 \theta }}{6 \cos ^2 \theta}$

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Precalculus : Find the Polar Equation of a Conic Section

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Example Question #11 : Find The Polar Equation Of A Conic Section

Example Question #41 : Conic Sections

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