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Precalculus : Polar Coordinates

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Example Question #1 : Polar Coordinates

Convert the polar coordinates to rectangular coordinates:

$\left(5, \frac{3\pi}{2}\right)$

Possible Answers:

(5, 0)

(0, -5)

(-5, -5)

(-5, 0)

Correct answer:

(0, -5)

Explanation:

To convert polar coordinates $(r, \theta)$ to rectangular coordinates (x, y) ,

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=5\cos \frac{3\pi}{2}=5(0)=0$

$y=5 \sin \frac{3\pi}{2}=5(-1)=-5$

The rectangular coordinates are (0, -5)

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert from polar form to rectangular form:

$r=4\cos\theta$

Possible Answers:

x^2+y^2+4=0

(x+y)^2=4

(x-2)^2+y^2=4

(x+2)^2+y^2=4

Correct answer:

(x-2)^2+y^2=4

Explanation:

Start by multiplying both sides by .

$r=4\cos\theta$

$r^2=4r\cos\theta$

Keep in mind that r^2=x^2+y^2

$x^2+y^2=4r\cos\theta$

Remember that $x=r\cos\theta$

So then,

x^2+y^2=4x

x^2-4x+y^2=0

Now, complete the square.

(x^2-4x+4)+y^2=4

(x-2)^2+y^2=4

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Example Question #3 : Polar Coordinates

Convert the polar equation to rectangular form:

$r=8\sin\theta$

Possible Answers:

x^2+(y+4)^2=16

x^2+(y-4)^2=16

(x-4)^2+y^2=15

(x+y-4)^2=16

Correct answer:

x^2+(y-4)^2=16

Explanation:

Start by multiplying both sides by .

$r=8\sin\theta$

$r^2=8r\sin\theta$

Remember that r^2=x^2+y^2

$x^2+y^2=8r\sin\theta$

Keep in mind that $y=r\sin\theta$

So then,

x^2+y^2=8y

x^2+y^2-8x=0

Now, complete the square.

x^2+(y^2-8x+16)=16

x^2+(y-4)^2=16

This is a graph of a circle with a radius of and a center at (0, 4)

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Example Question #4 : Polar Coordinates

Convert the polar equation into rectangular form:

$r=8\sec\theta$

Possible Answers:

x^2+y^2=8

(x-1)^2+y^2=64

y=8

x=8

Correct answer:

x=8

Explanation:

Remember that $\sec\theta=\frac{1}{\cos\theta}$

So then $r=8\sec\theta$ becomes

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

Now, multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=8$

Since $x=r\cos\theta,$ the rectangular form of this equation is

x=8

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r^2=4\sin(2\theta)$

Possible Answers:

(x^2+y^2+8)=16

(x^2+y^2)^2=8xy

(x^2+y^2)^2=8

(x^2-^2)^2=8xy

Correct answer:

(x^2+y^2)^2=8xy

Explanation:

Start by using the double angle formula for $\sin$ .

$\sin(2\theta)=2\sin\theta\cos\theta$

Substitute that into the equation gives the following:

$r^2=4(2\sin\theta\cos\theta)$

$r^2=8\sin\theta\cos\theta$

Because we need $r\cos\theta$ and $r\sin\theta$ to get and respectively, multiply both sides by r^2 .

$(r^2)(r^2)=8(r\sin\theta)(r\cos\theta)$

Now, recall that r^2=x^2+y^2 .

(x^2+y^2)(x^2+y^2)=8xy

(x^2+y^2)^2=8xy

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Example Question #6 : Polar Coordinates

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{6}$

Possible Answers:

$y=-\frac{\sqrt3}{3}x$

$y=-\sqrt3x$

$y=-\frac{3}\sqrt3}x$

y=-3x

Correct answer:

$y=-\frac{\sqrt3}{3}x$

Explanation:

Start by taking the tangent.

$\tan\theta=\tan\frac{5\pi}{6}$

$\tan\theta=-\frac{\sqrt3}{3}$

Recall that $\tan\theta=\frac{y}{x}$

$\frac{y}{x}=-\frac{\sqrt3}{3}$

$y=-\frac{\sqrt3}{3}x$

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Example Question #7 : Polar Coordinates

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{4}$

Possible Answers:

$y=\frac{5\pi}{4}$

y=-x

y=x

$x=\frac{5\pi}{4}$

Correct answer:

y=x

Explanation:

Start by taking the tangent.

$\tan\theta=\tan\frac{5\pi}{4}$

$\tan\theta=1$

Recall that $\tan\theta=\frac{y}{x}$

$\frac{y}{x}=1$

y=x

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation to rectangular form:

$r=4\sec\theta\tan\theta$

Possible Answers:

$y=2\sqrt x$

$y=\frac{1}{2}x^2$

x=4y^2

$y=\frac{1}{4}x^2$

Correct answer:

$y=\frac{1}{4}x^2$

Explanation:

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Substituting this into the given equation gives

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

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=4\tan\theta$

Recall that $x=r\cos\theta$ and that $\tan\theta=\frac{y}{x}$

$x=4\frac{y}{x}$

4y=x^2

$y=\frac{1}{4}x^2$

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Example Question #9 : Polar Coordinates

Convert the polar equation into rectangular form:

$r=2\csc\theta\cot\theta$

Possible Answers:

$y=\sqrt{2x}$

$y=\sqrt{\frac{x}{2}}$

$y=-\sqrt{2x}$

y=2x^2

Correct answer:

$y=\sqrt{2x}$

Explanation:

Recall that $\csc\theta=\frac{1}{sin\theta}$ and $\cot\theta=\frac{1}{tan\theta}$

$r=2\frac{1}{\sin\theta}\frac{1}{\tan\theta}$

Multiply both sides by $\sin\theta$

$r\sin\theta=2(\frac{1}{\tan\theta})$

Recall that $\tan\theta=\frac{y}{x}$ and $r\sin\theta=y$

$y=2(\frac{1}{\frac{y}{x}})$

$y=2(\frac{x}{y})$

2x=y^2

$y=\sqrt{2x}$

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=-3\sec\theta$

Possible Answers:

y=-3

x^2+y^2=-3

x=-3

(x-3)^2=4

Correct answer:

x=-3

Explanation:

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Plugging this into the equation gives us

$r=-3\frac{1}{\cos\theta}$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=-3$

Recall that $r\cos\theta=x$

So then the rectangular form of the equation is x=-3

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Precalculus : Polar Coordinates

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Polar Coordinates

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #3 : Polar Coordinates

Example Question #4 : Polar Coordinates

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #6 : Polar Coordinates

Example Question #7 : Polar Coordinates

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #9 : Polar Coordinates

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

All Precalculus Resources

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