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 [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.
- Determine the sign in the denominator. If [latex]p<0[/latex], use subtraction. If [latex]p>0[/latex], use addition.
- Write the coefficient of the trigonometric function as the given eccentricity.
- 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
and [latex]e=3[/latex] and [latex]|-2|=2=p[/latex].
Therefore,
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
and [latex]e=\frac{3}{5}[/latex] and [latex]|4|=4=p[/latex].
Therefore,
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].
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].
Try It 4
Convert the conic [latex]r=\frac{2}{1+2\text{ }\cos \text{ }\theta }[/latex] to rectangular form.