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Precalculus : Polar Equations of Conic Sections

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Example Questions

Example Question #6 : Find The Polar Equation Of A Conic Section

Find the polar equation for $\frac{x^2 }{25} + \frac{ (y - \sqrt{10} ) ^2 }{100 } = 1$ .

Possible Answers:

$r = \frac{ \sqrt{10 }}{ \sqrt{20 } + \cos \theta}$

$r = \frac{ 10} { \sqrt{20 } - \sqrt{10} \cos \theta }$

$r = \frac{ 10 }{ \sqrt{20 } + \sqrt{10 } \sin \theta }$

$r = \frac{25} { \sqrt{10 } - 5\sqrt{3 } \sin \theta }$

Correct answer:

$r = \frac{25} { \sqrt{10 } - 5\sqrt{3 } \sin \theta }$

Explanation:

To put this in polar form, we need to understand its structure. We can find the foci by using the relationship b^2 = a^2 - c^2 where a is half the length of the major axis, and b is half the length of the minor axis.

In this problem,

a=10, b=5 .

Plugging these values into the above realationship we can solve for .

5^2 = 10^2 - c^ 2

-75 = -c^2 divide by -1

75 = c^2

$\sqrt{75 } =5\sqrt{3}= c$

Since the center is $(0, 5\sqrt{3})$ , the foci are at (0, 0 ) and $(0, 10\sqrt{3})$ .

The major axis is vertical, and the lower focus is at the origin, so the polar equation will be in the form $r = \frac{pe }{ 1 - e \cdot \sin \theta }$ where $e = \frac{ c}{a}$ and for an ellipse, $p = \frac{a^2 - c^2 } { c }$ .

Here, $e = \frac{ 5\sqrt {3}}{ \sqrt{10}}$ and $p = \frac{ 100 - 75 }{ 5\sqrt{3} } = \frac{ 5}{\sqrt{3}}$

$r = \frac{ \frac{5 }{\sqrt{3 } } \cdot \frac{ 5\sqrt{3}}{\sqrt{10}}}{1 - \frac{5\sqrt{3}}{\sqrt{10} } \sin \theta}$

Simplify the top by cancelling out $\sqrt{3}$

$r = \frac{ \frac{25}{\sqrt{10}}}{ 1 - \frac{ 5\sqrt{3}}{\sqrt{10}} \sin \theta}$ Simplify by multiplying top and bottom by $\sqrt{10}$

$r = \frac{25} { \sqrt{10 } - 5\sqrt{3 } \sin \theta }$

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Example Question #31 : Conic Sections

Write the equation $\frac{ (x-6)^2 }{27} - \frac{y^2 }{9 } = 1$ in polar form.

Possible Answers:

$r = \frac{ 3 } { 1 + 9 \cos \theta }$

$r = \frac{ 3} { \sqrt3 + 2 \sin \theta }$

$r = \frac{ 6 }{ 2 - \sqrt{3} \cos \theta }$

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

Correct answer:

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

Explanation:

Before writing this in polar form, we need to know the structure of this hyperbola. We can determine the foci by using the relationship b^2 = c^2 - a^2

In this case, 9 = 27 - c^2 subtract 27

-36 = -c^2

36 = c^2

6 = c

The center is at (6, 0) and the major axis is horizontal, so the foci are (0,0) and (12, 0 ) , adding or subtracting 6 from the x-coordinate. Because the left focus is at the origin and the major axis is horizontal, our polar equation will be in the form $r = \frac{pe}{ 1 + e \cdot \cos \theta }$ where $e = \frac{ c}{a }$ and for a hyperbola $p = \frac{c^2 - a^2 }{c }$

Here, $e = \frac{ 6}{ \sqrt{27 } } = \frac{ 6 }{ 3 \sqrt{3} } = \frac{ 2}{ \sqrt3 }$

$p = \frac{ 36 - 27 }{ 6 } = \frac{9 }{6} = \frac{ 3}{2}$

Plugging this in gives:

$r = \frac{ \frac{ 3}{2 } \cdot \frac{ 2}{ \sqrt 3 } }{ 1 + \frac{2 }{ \sqrt{3} } \cos \theta }$ simplify the numerator by cancelling the 2's

$r = \frac{ \frac{ 3}{ \sqrt 3 }}{ 1 + \frac{ 2} {\sqrt3} \cos \theta }$ simplify once more by multiplying top and bottom by $\sqrt3$

$r = \frac{ 3}{ \sqrt3 + 2 \cos \theta }$

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Example Question #1 : Find The Polar Equation Of A Conic Section

Write the polar equation equivalent to 16(x+ 4 ) = y^2 .

Possible Answers:

$r = \frac{ 8 }{ 1 + \cos \theta }$

$r = \frac{ 4}{ 1 + \sin \theta }$

$r = \frac{ 4 }{ 1 - \cos \theta }$

$r = \frac{ 8 }{ 1 - \sin \theta }$

$r = \frac{ 8 }{ 1 - \cos \theta }$

Correct answer:

$r = \frac{ 8 }{ 1 - \cos \theta }$

Explanation:

This is the equation for a right-opening parabola with a vertex at (-4, 0 ) . The distance from the vertex to the focus can be found by solving 4p = 16 , so p = 4 . This places the focus at (0,0) .

Because the focus is at the origin, and the parabola opens to the right, this equation is in the form $r = \frac{ 2p }{ 1 - \cos \theta }$ .

This particular parabola has a polar equation $r = \frac{ 8 }{ 1 - \cos \theta}$ .

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Example Question #5 : Find The Polar Equation Of A Conic Section

Write the equation for -20 (y- 5 ) = x ^2 in polar form.

Possible Answers:

$r = \frac{ 10 }{ 1 - \sin \theta }$

$r = \frac{ -10 }{ 1 - \sin \theta }$

$r = \frac{ 10 }{ 1 - \cos \theta }$

$r = \frac{ 10 }{ 1 + \sin \theta}$

$r = \frac{ -10}{ 1 + \sin \theta}$

Correct answer:

$r = \frac{ 10 }{ 1 + \sin \theta}$

Explanation:

This is the equation for a down-opening parabola. The vertex is at (0, 5) . We can figure out the location of the focus by solving 4 p = 20 . This means that p = 5 , so the focus is at (0,0) .

Because the focus is at the origin, and the parabola opens down, the polar form of the equation is $r = \frac{ 2p }{ 1 + \sin \theta }$ .

This equation is

$r = \frac{ 10 }{ 1 + \sin \theta }$ .

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Example Question #10 : Find The Polar Equation Of A Conic Section

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

Possible Answers:

$r = -7 + 6 \cos \theta$

$r=3 \cos \theta + \sqrt{9 \cos ^2 \theta -7}$

$r = 6 \cos \theta - \sqrt{36 \cos ^2 \theta - 28 }$

$r = 6 \cos \theta - \sqrt{36\cos ^2 \theta + 8 }$

$r = \sqrt{9\cos^2 \theta +2 }$

Correct answer:

$r=3 \cos \theta + \sqrt{9 \cos ^2 \theta -7}$

Explanation:

First, multiply out (x-3)^2 :

x^2 -6x + 9 + y^2 = 2 we can re-arrange this a little bit by subtracting 2 from both sides and putting x^2 next to y^2 :

x^2 + y^2 -6x +7 = 0

Now we can make the substitutions x^2 + y^2 = r^2 and $x = r \cos \theta$ :

$r^2 -6 r \cos \theta + 7 = 0$

We can solve for r using the quadratic formula:

$r = \frac{ 6 \cos \theta \pm \sqrt{36\cos^2\theta - 4(1)(7)}}{2(1)}$

$r = \frac{6 \cos \theta \pm \sqrt{36 \cos ^2 \theta - 28}}{2}$ factor out 4 inside the square root

$r = \frac{6 \cos \theta \pm \sqrt{4(9\cos^2 \theta - 7 )}}{2 }$

$r = \frac{6 \cos \theta \pm \sqrt4\sqrt{9\cos^2 \theta - 7 }}{2 } = \frac{6 \cos \theta \pm 2 \sqrt{9\cos ^2 -7}}{2}$

$r = 3 \cos \theta \pm \sqrt{9\cos^2 \theta - 7 }$

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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 : Polar Equations of Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #6 : Find The Polar Equation Of A Conic Section

Example Question #31 : Conic Sections

Example Question #1 : Find The Polar Equation Of A Conic Section

Example Question #5 : Find The Polar Equation Of A Conic Section

Example Question #10 : Find The Polar Equation Of A Conic Section

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

Example Question #41 : Conic Sections

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