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.

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.
Determine the sign in the denominator. If $p<0$ , use subtraction. If $p>0$ , use addition.
Write the coefficient of the trigonometric function as the given eccentricity.
Write the absolute value of $p$ 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, $e=3$ and directrix $y=-2$ .

Solution

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

$r=\frac{ep}{1-e\text{ }\sin \text{ }\theta }$

and $e=3$ and $|-2|=2=p$ .

Therefore,

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

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, $e=\frac{3}{5}$ , and directrix $x=4$ .

Solution

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

$r=\frac{ep}{1+e\text{ }\cos \text{ }\theta }$

and $e=\frac{3}{5}$ and $|4|=4=p$ .

Therefore,

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

Try It 3

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

Solution

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

Convert the conic $r=\frac{1}{5 - 5\sin \theta }$ to rectangular form.

Solution

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

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

Try It 4

Convert the conic $r=\frac{2}{1+2\text{ }\cos \text{ }\theta }$ to rectangular form.

Solution

Conic Sections in Polar Coordinates