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.

How To: 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 [latex]y[/latex], we use the general polar form in terms of sine. If the directrix is given in terms of [latex]x[/latex], we use the general polar form in terms of cosine.
  2. Determine the sign in the denominator. If [latex]p<0[/latex], use subtraction. If [latex]p>0[/latex], use addition.
  3. Write the coefficient of the trigonometric function as the given eccentricity.
  4. Write the absolute value of [latex]p[/latex] in the numerator, and simplify the equation.

Example 5: 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, [latex]e=3[/latex] and directrix [latex]y=-2[/latex].

Solution

The directrix is [latex]y=-p[/latex], so we know the trigonometric function in the denominator is sine.

Because [latex]y=-2,-2<0[/latex], so we know there is a subtraction sign in the denominator. We use the standard form of

[latex]r=\frac{ep}{1-e\text{ }\sin \text{ }\theta }[/latex]

and [latex]e=3[/latex] and [latex]|-2|=2=p[/latex].

Therefore,

[latex]\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{\left(3\right)\left(2\right)}{1 - 3\text{ }\sin \text{ }\theta }\hfill \\ r=\frac{6}{1 - 3\text{ }\sin \text{ }\theta }\hfill \end{array}\hfill \end{array}[/latex]

Example 6: 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, [latex]e=\frac{3}{5}[/latex], and directrix [latex]x=4[/latex].

Solution

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

[latex]r=\frac{ep}{1+e\text{ }\cos \text{ }\theta }[/latex]

and [latex]e=\frac{3}{5}[/latex] and [latex]|4|=4=p[/latex].

Therefore,

[latex]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{\left(\frac{3}{5}\right)\left(4\right)}{1+\frac{3}{5}\cos \theta }\hfill \end{array}\hfill \\ r=\frac{\frac{12}{5}}{1+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{\frac{12}{5}}{1\left(\frac{5}{5}\right)+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{12}{5}\cdot \frac{5}{5+3\cos \theta }\hfill \\ r=\frac{12}{5+3\cos \theta }\hfill \end{array}[/latex]

Try It 3

Find the polar form of the conic given a focus at the origin, [latex]e=1[/latex], and directrix [latex]x=-1[/latex].

Solution

Example 5: Converting a Conic in Polar Form to Rectangular Form

Convert the conic [latex]r=\frac{1}{5 - 5\sin \theta }[/latex] to rectangular form.

Solution

We will rearrange the formula to use the identities [latex]r=\sqrt{{x}^{2}+{y}^{2}},x=r\cos \theta ,\text{and }y=r\sin \theta[/latex].

[latex]\begin{array}{ll}\text{ }r=\frac{1}{5 - 5\sin \theta }\hfill & \hfill \\ r\cdot \left(5 - 5\sin \theta \right)=\frac{1}{5 - 5\sin \theta }\cdot \left(5 - 5\sin \theta \right)\hfill & \text{Eliminate the fraction}.\hfill \\ \text{ }5r - 5r\sin \theta =1\hfill & \text{Distribute}.\hfill \\ \text{ }5r=1+5r\sin \theta \hfill & \text{Isolate }5r.\hfill \\ \text{ }25{r}^{2}={\left(1+5r\sin \theta \right)}^{2}\hfill & \text{Square both sides}.\hfill \\ \text{ }25\left({x}^{2}+{y}^{2}\right)={\left(1+5y\right)}^{2}\hfill & \text{Substitute }r=\sqrt{{x}^{2}+{y}^{2}}\text{ and }y=r\sin \theta .\hfill \\ \text{ }25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}\hfill & \text{Distribute and use FOIL}.\hfill \\ \text{ }25{x}^{2}-10y=1\hfill & \text{Rearrange terms and set equal to 1}.\hfill \end{array}[/latex]

Try It 4

Convert the conic [latex]r=\frac{2}{1+2\text{ }\cos \text{ }\theta }[/latex] to rectangular form.

Solution