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Graphing a hyperbola in polar form

Graph r = 8 2 3   sin   θ .

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 1 2 .

r = 8 2 3 sin   θ = 8 ( 1 2 ) 2 ( 1 2 ) 3 ( 1 2 ) sin   θ r = 4 1 3 2   sin   θ

Because e = 3 2 , e > 1 , so we will graph a hyperbola    with a focus at the origin. The function has a sin   θ term and there is a subtraction sign in the denominator, so the directrix is y = p .

       4 = e p        4 = ( 3 2 ) p 4 ( 2 3 ) = p       8 3 = p

The directrix is y = 8 3 .

Plotting a few key points as in [link] will enable us to see the vertices. See [link] .

A B C D
θ 0 π 2 π 3 π 2
r = 8 2 3 sin θ 4 8 4 8 5 = 1.6
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Graphing an ellipse in polar form

Graph r = 10 5 4   cos   θ .

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 1 5 .

r = 10 5 4 cos   θ = 10 ( 1 5 ) 5 ( 1 5 ) 4 ( 1 5 ) cos   θ r = 2 1 4 5   cos   θ

Because e = 4 5 , e < 1 , so we will graph an ellipse    with a focus at the origin. The function has a cos θ , and there is a subtraction sign in the denominator, so the directrix    is x = p .

       2 = e p        2 = ( 4 5 ) p 2 ( 5 4 ) = p       5 2 = p

The directrix is x = 5 2 .

Plotting a few key points as in [link] will enable us to see the vertices. See [link] .

A B C D
θ 0 π 2 π 3 π 2
r = 10 5 4   cos   θ 10 2 10 9 1.1 2
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Graph r = 2 4 cos   θ .

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Defining conics in terms of a focus and a directrix

So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.

Given the focus, eccentricity, and directrix of a conic, determine the polar equation.

  1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y , we use the general polar form in terms of sine. If the directrix is given in terms of x , we use the general polar form in terms of cosine.
  2. Determine the sign in the denominator. If p < 0 , use subtraction. If p > 0 , use addition.
  3. Write the coefficient of the trigonometric function as the given eccentricity.
  4. Write the absolute value of p in the numerator, and simplify the equation.

Finding the polar form of a vertical conic given a focus at the origin and the eccentricity and directrix

Find the polar form of the conic given a focus at the origin, e = 3 and directrix     y = 2.

The directrix is y = p , so we know the trigonometric function in the denominator is sine.

Because y = −2 , –2 < 0 , so we know there is a subtraction sign in the denominator. We use the standard form of

r = e p 1 e   sin   θ

and e = 3 and | −2 | = 2 = p .

Therefore,

r = ( 3 ) ( 2 ) 1 3   sin   θ r = 6 1 3   sin   θ
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Finding the polar form of a horizontal conic given a focus at the origin and the eccentricity and directrix

Find the polar form of a conic given a focus at the origin, e = 3 5 , and directrix     x = 4.

Because the directrix is x = p , we know the function in the denominator is cosine. Because x = 4 , 4 > 0 , so we know there is an addition sign in the denominator. We use the standard form of

r = e p 1 + e   cos   θ

and e = 3 5 and | 4 | = 4 = p .

Therefore,

r = ( 3 5 ) ( 4 ) 1 + 3 5 cos θ r = 12 5 1 + 3 5 cos θ r = 12 5 1 ( 5 5 ) + 3 5 cos θ r = 12 5 5 5 + 3 5 cos θ r = 12 5 5 5 + 3 cos θ r = 12 5 + 3 cos θ
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Find the polar form of the conic given a focus at the origin, e = 1 , and directrix x = −1.

r = 1 1 cos θ

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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