Find the Polar Equation of a Conic Section - Pre-Calculus

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Question

Which is the correct polar form of the cartesian equation $\frac{(y+1)^2 }{ 4 } + \frac{x^2 }{3 } = 1$ ?

Answer

To determine the polar equation, first we need to interpret the original cartesian graph. This is an ellipse with a vertical major axis with half its length $a = \sqrt{4} = 2$ . The minor axis has half its length $b = \sqrt{3}$ . To find the foci, use the relationship

$\sqrt{3} ^ 2 = 2^ 2 - c ^ 2$

so

Since the center is , this means that the foci are at $(0, -1 \pm 1 ) = (0, -2); (0, 0 )$

Having a focus at the origin means we can use the formula $r = \frac{pe}{ 1 \pm e \cos \theta}$ \[or sine\] where e is the eccentricity, and for an ellipse $p = \frac{a^2 - c^2 }{ c }$ .

In this case, the focus at the origin is above the directrix, so we be subtracting. The major axis is vertical, so we are using sine.

Solving for p gives us: $p = \frac{2^2 - 1^2 }{ 1 } = 4 - 1 = 3$

The eccentricity is $e = \frac {c }{a }$ , in this case $\frac{1}{2 }$

This gives us an equation of:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta}$

We can simplify by multiplying top and bottom by 2:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta} * \frac{2}{2 } = \frac{ 3 }{ 2 - \sin \theta}$

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Question

Write the equation for the circle in polar form.

Answer

To convert this cartesian equation to polar form, we will use the substitutions and $x = r \cos \theta$ .

First, we should expand the expression:

square x - 3

subtract 9 from both sides

group the squared variables next to each other - this will help see how to re-write this in polar form:

now make the substitutions:

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

This is a quadratic in r with $a = 1, b = -6 \cos \theta, c = 0$

Solve using the quadratic formula:

$r = \frac {6 \cos \theta \pm \sqrt {(6 \cos \theta)^2 - 410}}{2*1} = \frac{6 \cos \theta \pm \sqrt{ (6 \cos \theta )^2 }}{2} = \frac {6 \cos \theta \pm 6 \cos \theta }{2}$

Subtracting $6 \cos \theta - 6 \cos \theta = 0$ which would give us a circle with no radius, but adding $6 \cos \theta + 6 \cos \theta = 12 \cos \theta$ so our answer is:

$r = \frac{12 \cos \theta }{2 } = 6 \cos \theta$

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Question

Write the equation for the hyperbola $\frac{(x + \sqrt {11} ) ^ 2 }{9 } - \frac{y^2 }{2} = 1$ in polar form. Note that the rightmost focus is at the origin \[directrix is to the left\].

Answer

The directrix is to the left of the focus at the origin, and the major axis is horizontal, so our equation is going to take the form

$r = \frac{pe}{ 1 - e \cos \theta }$ where e is the eccentricity. For a hyperbola specifically, $p = \frac{c^2 - a^2 }{c }$

First we need to solve for c so we can find the eccentricity. For a hyperbola, we use the relationship where is half the length of the minor axis and is half the length of the major axis, in this case the horizontal one.

In this case, $b^2 = 2; b = \sqrt{2}$ and

add 9 to both sides

take the square root

$\sqrt{11} = c$

To find the eccentricity: $e = \frac{c}{a}$ so in this case $e = \frac{\sqrt{11}}{3}$

$p = \frac{c^2 - a^2 }{c}$ gives us $p = \frac{11 - 9 }{\sqrt{11}} = \frac{2}{\sqrt{11}}$

Plugging in e and p:

$r = \frac{\frac{2}{\sqrt{11}} * \frac{\sqrt{11}}{3}}{ 1 + \frac{\sqrt{11}}{3 } \cos \theta} = \frac{\frac{2}{3}}{1 + \frac{\sqrt{11}}3 \cos \theta}$

To simplify we can multiply top and bottom by 3:

$r = \frac{2}{3 + \sqrt{11} \cos \theta}$

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Question

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

Answer

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

In this problem,

.

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

divide by -1

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

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

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

Answer

Before writing this in polar form, we need to know the structure of this hyperbola. We can determine the foci by using the relationship

In this case, subtract 27

The center is at and the major axis is horizontal, so the foci are and , 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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Question

Write the equation for in polar form.

Answer

This is the equation for a parabola, so the eccentricity is 1. It opens up, so the focus is above the directrix.

This means that our equation will be in the form

$r= \frac{ 2a}{ 1 - \sin \theta }$

where a is the distance from the focus to the vertex.

In this case, we have , so .

Because the vertex is , the focus is so we can use the formula without adjusting anything.

The equation is

$r = \frac{ 4 }{ 1 - \sin \theta}$ .

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Question

Write the equation for in polar form.

Answer

This is the equation for a parabola with a vertex at . This parabola opens left because it is negative. The distance from the focus to the vertex can be found by solving , so . This places the focus at the origin.

Because of this and the fact that the focus is to the left of the directrix, we know that the polar equation is in the form

$r = \frac{ 2a }{ 1 + \cos \theta }$ .

That means that this equation is

$r = \frac{ 2 }{ 1 + \cos \theta}$ .

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Question

Write the polar equation equivalent to .

Answer

This is the equation for a right-opening parabola with a vertex at . The distance from the vertex to the focus can be found by solving , so . This places the focus at .

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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Question

Write the equation for in polar form.

Answer

This is the equation for a down-opening parabola. The vertex is at . We can figure out the location of the focus by solving . This means that , so the focus is at .

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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Question

Write the equation in polar form

Answer

First, multiply out :

we can re-arrange this a little bit by subtracting 2 from both sides and putting next to :

Now we can make the substitutions 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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Question

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

Answer

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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Question

Write the equation in polar form

Answer

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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Question

Which is the correct polar form of the cartesian equation $\frac{(y+1)^2 }{ 4 } + \frac{x^2 }{3 } = 1$ ?

Answer

To determine the polar equation, first we need to interpret the original cartesian graph. This is an ellipse with a vertical major axis with half its length $a = \sqrt{4} = 2$ . The minor axis has half its length $b = \sqrt{3}$ . To find the foci, use the relationship

$\sqrt{3} ^ 2 = 2^ 2 - c ^ 2$

so

Since the center is , this means that the foci are at $(0, -1 \pm 1 ) = (0, -2); (0, 0 )$

Having a focus at the origin means we can use the formula $r = \frac{pe}{ 1 \pm e \cos \theta}$ \[or sine\] where e is the eccentricity, and for an ellipse $p = \frac{a^2 - c^2 }{ c }$ .

In this case, the focus at the origin is above the directrix, so we be subtracting. The major axis is vertical, so we are using sine.

Solving for p gives us: $p = \frac{2^2 - 1^2 }{ 1 } = 4 - 1 = 3$

The eccentricity is $e = \frac {c }{a }$ , in this case $\frac{1}{2 }$

This gives us an equation of:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta}$

We can simplify by multiplying top and bottom by 2:

$r = \frac{3\frac{1}{2}}{ 1 - \frac{1}{2} \sin \theta} * \frac{2}{2 } = \frac{ 3 }{ 2 - \sin \theta}$

Compare your answer with the correct one above

Question

Write the equation for the circle in polar form.

Answer

To convert this cartesian equation to polar form, we will use the substitutions and $x = r \cos \theta$ .

First, we should expand the expression:

square x - 3

subtract 9 from both sides

group the squared variables next to each other - this will help see how to re-write this in polar form:

now make the substitutions:

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

This is a quadratic in r with $a = 1, b = -6 \cos \theta, c = 0$

Solve using the quadratic formula:

$r = \frac {6 \cos \theta \pm \sqrt {(6 \cos \theta)^2 - 410}}{2*1} = \frac{6 \cos \theta \pm \sqrt{ (6 \cos \theta )^2 }}{2} = \frac {6 \cos \theta \pm 6 \cos \theta }{2}$

Subtracting $6 \cos \theta - 6 \cos \theta = 0$ which would give us a circle with no radius, but adding $6 \cos \theta + 6 \cos \theta = 12 \cos \theta$ so our answer is:

$r = \frac{12 \cos \theta }{2 } = 6 \cos \theta$

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Question

Write the equation for the hyperbola $\frac{(x + \sqrt {11} ) ^ 2 }{9 } - \frac{y^2 }{2} = 1$ in polar form. Note that the rightmost focus is at the origin \[directrix is to the left\].

Answer

The directrix is to the left of the focus at the origin, and the major axis is horizontal, so our equation is going to take the form

$r = \frac{pe}{ 1 - e \cos \theta }$ where e is the eccentricity. For a hyperbola specifically, $p = \frac{c^2 - a^2 }{c }$

First we need to solve for c so we can find the eccentricity. For a hyperbola, we use the relationship where is half the length of the minor axis and is half the length of the major axis, in this case the horizontal one.

In this case, $b^2 = 2; b = \sqrt{2}$ and

add 9 to both sides

take the square root

$\sqrt{11} = c$

To find the eccentricity: $e = \frac{c}{a}$ so in this case $e = \frac{\sqrt{11}}{3}$

$p = \frac{c^2 - a^2 }{c}$ gives us $p = \frac{11 - 9 }{\sqrt{11}} = \frac{2}{\sqrt{11}}$

Plugging in e and p:

$r = \frac{\frac{2}{\sqrt{11}} * \frac{\sqrt{11}}{3}}{ 1 + \frac{\sqrt{11}}{3 } \cos \theta} = \frac{\frac{2}{3}}{1 + \frac{\sqrt{11}}3 \cos \theta}$

To simplify we can multiply top and bottom by 3:

$r = \frac{2}{3 + \sqrt{11} \cos \theta}$

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Question

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

Answer

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

In this problem,

.

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

divide by -1

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

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

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

Answer

Before writing this in polar form, we need to know the structure of this hyperbola. We can determine the foci by using the relationship

In this case, subtract 27

The center is at and the major axis is horizontal, so the foci are and , 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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Question

Write the equation for in polar form.

Answer

This is the equation for a parabola, so the eccentricity is 1. It opens up, so the focus is above the directrix.

This means that our equation will be in the form

$r= \frac{ 2a}{ 1 - \sin \theta }$

where a is the distance from the focus to the vertex.

In this case, we have , so .

Because the vertex is , the focus is so we can use the formula without adjusting anything.

The equation is

$r = \frac{ 4 }{ 1 - \sin \theta}$ .

Compare your answer with the correct one above

Question

Write the equation for in polar form.

Answer

This is the equation for a parabola with a vertex at . This parabola opens left because it is negative. The distance from the focus to the vertex can be found by solving , so . This places the focus at the origin.

Because of this and the fact that the focus is to the left of the directrix, we know that the polar equation is in the form

$r = \frac{ 2a }{ 1 + \cos \theta }$ .

That means that this equation is

$r = \frac{ 2 }{ 1 + \cos \theta}$ .

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Question

Write the polar equation equivalent to .

Answer

This is the equation for a right-opening parabola with a vertex at . The distance from the vertex to the focus can be found by solving , so . This places the focus at .

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

Compare your answer with the correct one above

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