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Exercises Conic sections

Example 5 from Algebra and Trigonometry, 12.5 Conic Sections in Polar Coordinates

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 $y = - 2.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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

$r = \frac{e p}{1 - e \sin}$

and $e = 3$ and $| −2 | = 2 = p .$

Therefore,

$\begin{array}{l} \begin{array}{l} r = \frac{(3) (2)}{1 - 3 \sin} \\ r = \frac{6}{1 - 3 \sin} \end{array} \end{array}$

How did it go?

PreviousGraphing an Ellipse in Polar Form Next Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 12.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.