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Study Guides > College Algebra

Identifying a Conic in Polar Form

Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x=2+y2x=2+{y}^{2} shown in Figure 2.

Figure 2
In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r,θ)P\left(r,\theta \right) at the pole, and a line, the directrix, which is perpendicular to the polar axis. If FF is a fixed point, the focus, and DD is a fixed line, the directrix, then we can let ee be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points PP such that e=PFPDe=\frac{PF}{PD} is a conic. In other words, we can define a conic as the set of all points PP with the property that the ratio of the distance from PP to FF to the distance from PP to DD is equal to the constant ee. For a conic with eccentricity ee,

  • if 0e<10\le e<1, the conic is an ellipse
  • if e=1e=1, the conic is a parabola
  • if e>1e>1, the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, x=±px=\pm p, the eccentricity ee, and the angle θ\theta . Thus, each conic may be written as a polar equation, an equation written in terms of rr and θ\theta .

A General Note: The Polar Equation for a Conic

For a conic with a focus at the origin, if the directrix is x=±px=\pm p, where pp is a positive real number, and the eccentricity is a positive real number ee, the conic has a polar equation
r=ep1±e cos θr=\frac{ep}{1\pm e\text{ }\cos \text{ }\theta }
For a conic with a focus at the origin, if the directrix is y=±py=\pm p, where pp is a positive real number, and the eccentricity is a positive real number ee, the conic has a polar equation
r=ep1±e sin θr=\frac{ep}{1\pm e\text{ }\sin \text{ }\theta }

How To: 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 ee as the coefficient of the trigonometric function in the denominator.
  3. Compare ee with 1 to determine the shape of the conic.
  4. Determine the directrix as x=px=p if cosine is in the denominator and y=py=p if sine is in the denominator. Set epep equal to the numerator in standard form to solve for xx or yy.

Example 1: 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=63+2 sin θr=\frac{6}{3+2\text{ }\sin \text{ }\theta }
  2. r=124+5 cos θr=\frac{12}{4+5\text{ }\cos \text{ }\theta }
  3. r=722 sin θr=\frac{7}{2 - 2\text{ }\sin \text{ }\theta }

Solution

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, 1c\frac{1}{c}, where cc is that constant.
  1. Multiply the numerator and denominator by 13\frac{1}{3}.
    r=63+2sin θ(13)(13)=6(13)3(13)+2(13)sin θ=21+23 sin θr=\frac{6}{3+2\sin \text{ }\theta }\cdot \frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{6\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)\sin \text{ }\theta }=\frac{2}{1+\frac{2}{3}\text{ }\sin \text{ }\theta }
    Because sin θ\sin \text{ }\theta is in the denominator, the directrix is y=py=p. Comparing to standard form, note that e=23e=\frac{2}{3}. Therefore, from the numerator,
     2=ep 2=23p(32)2=(32)23p 3=p\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\frac{2}{3}p\hfill \\ \left(\frac{3}{2}\right)2=\left(\frac{3}{2}\right)\frac{2}{3}p\hfill \\ \text{ }3=p\hfill \end{array}
    Since e<1e<1, the conic is an ellipse. The eccentricity is e=23e=\frac{2}{3} and the directrix is y=3y=3.
  2. Multiply the numerator and denominator by 14\frac{1}{4}.
    r=124+5 cos θ(14)(14)r=12(14)4(14)+5(14)cos θr=31+54 cos θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{12}{4+5\text{ }\cos \text{ }\theta }\cdot \frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)}\hfill \end{array}\hfill \\ r=\frac{12\left(\frac{1}{4}\right)}{4\left(\frac{1}{4}\right)+5\left(\frac{1}{4}\right)\cos \text{ }\theta }\hfill \\ r=\frac{3}{1+\frac{5}{4}\text{ }\cos \text{ }\theta }\hfill \end{array}
    Because  cosθ\text{ cos}\theta is in the denominator, the directrix is x=px=p. Comparing to standard form, e=54e=\frac{5}{4}. Therefore, from the numerator,
     3=ep 3=54p(45)3=(45)54p 125=p\begin{array}{l}\text{ }3=ep\hfill \\ \text{ }3=\frac{5}{4}p\hfill \\ \left(\frac{4}{5}\right)3=\left(\frac{4}{5}\right)\frac{5}{4}p\hfill \\ \text{ }\frac{12}{5}=p\hfill \end{array}
    Since e>1e>1, the conic is a hyperbola. The eccentricity is e=54e=\frac{5}{4} and the directrix is x=125=2.4x=\frac{12}{5}=2.4.
  3. Multiply the numerator and denominator by 12\frac{1}{2}.
    r=722 sin θ(12)(12)r=7(12)2(12)2(12) sin θr=721sin θ\begin{array}{l}\hfill \\ \hfill \\ \begin{array}{l}r=\frac{7}{2 - 2\text{ }\sin \text{ }\theta }\cdot \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}\hfill \\ r=\frac{7\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\text{ }\sin \text{ }\theta }\hfill \\ r=\frac{\frac{7}{2}}{1-\sin \text{ }\theta }\hfill \end{array}\hfill \end{array}
    Because sine is in the denominator, the directrix is y=py=-p. Comparing to standard form, e=1e=1. Therefore, from the numerator,
    72=ep72=(1)p72=p\begin{array}{l}\frac{7}{2}=ep\\ \frac{7}{2}=\left(1\right)p\\ \frac{7}{2}=p\end{array}
    Because e=1e=1, the conic is a parabola. The eccentricity is e=1e=1 and the directrix is y=72=3.5y=-\frac{7}{2}=-3.5.

Try It 1

Identify the conic with focus at the origin, the directrix, and the eccentricity for r=23cos θr=\frac{2}{3-\cos \text{ }\theta }. Solution

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