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Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.

  1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
  2. Identify the eccentricity e as the coefficient of the trigonometric function in the denominator.
  3. Compare e with 1 to determine the shape of the conic.
  4. Determine the directrix as x = p if cosine is in the denominator and y = p if sine is in the denominator. Set e p equal to the numerator in standard form to solve for x or y .

Identifying a conic given the polar form

For each of the following equations, identify the conic with focus at the origin, the directrix    , and the eccentricity    .

  1. r = 6 3 + 2   sin   θ
  2. r = 12 4 + 5   cos   θ
  3. r = 7 2 2   sin   θ

For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, 1 c , where c is that constant.

  1. Multiply the numerator and denominator by 1 3 .
    r = 6 3 + 2 sin   θ ( 1 3 ) ( 1 3 ) = 6 ( 1 3 ) 3 ( 1 3 ) + 2 ( 1 3 ) sin   θ = 2 1 + 2 3   sin   θ

    Because sin   θ is in the denominator, the directrix is y = p . Comparing to standard form, note that e = 2 3 . Therefore, from the numerator,

          2 = e p       2 = 2 3 p ( 3 2 ) 2 = ( 3 2 ) 2 3 p       3 = p

    Since e < 1 , the conic is an ellipse    . The eccentricity is e = 2 3 and the directrix is y = 3.

  2. Multiply the numerator and denominator by 1 4 .
    r = 12 4 + 5   cos   θ ( 1 4 ) ( 1 4 ) r = 12 ( 1 4 ) 4 ( 1 4 ) + 5 ( 1 4 ) cos   θ r = 3 1 + 5 4   cos   θ

    Because  cos θ   is in the denominator, the directrix is x = p . Comparing to standard form, e = 5 4 . Therefore, from the numerator,

           3 = e p        3 = 5 4 p ( 4 5 ) 3 = ( 4 5 ) 5 4 p     12 5 = p

    Since e > 1 , the conic is a hyperbola    . The eccentricity is e = 5 4 and the directrix is x = 12 5 = 2.4.

  3. Multiply the numerator and denominator by 1 2 .
    r = 7 2 2   sin   θ ( 1 2 ) ( 1 2 ) r = 7 ( 1 2 ) 2 ( 1 2 ) 2 ( 1 2 )   sin   θ r = 7 2 1 sin   θ

    Because sine is in the denominator, the directrix is y = p . Comparing to standard form, e = 1. Therefore, from the numerator,

    7 2 = e p 7 2 = ( 1 ) p 7 2 = p

    Because e = 1 , the conic is a parabola    . The eccentricity is e = 1 and the directrix is y = 7 2 = −3.5.

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Identify the conic with focus at the origin, the directrix, and the eccentricity for r = 2 3 cos   θ .

ellipse; e = 1 3 ; x = 2

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Graphing the polar equations of conics

When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e and, therefore, the shape of the curve. The next step is to substitute values for θ and solve for r to plot a few key points. Setting θ equal to 0 , π 2 , π , and 3 π 2 provides the vertices so we can create a rough sketch of the graph.

Graphing a parabola in polar form

Graph r = 5 3 + 3   cos   θ .

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 1 3 .

r = 5 3 + 3   cos   θ = 5 ( 1 3 ) 3 ( 1 3 ) + 3 ( 1 3 ) cos   θ r = 5 3 1 + cos   θ

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

5 3 = e p 5 3 = ( 1 ) p 5 3 = p

The directrix is x = 5 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 = 5 3 + 3   cos   θ 5 6 0.83 5 3 1.67 undefined 5 3 1.67
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Practice Key Terms 2

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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