# Thread: need help with polar equations.

1. ## need help with polar equations.

How do i create an equation for a parabola in a polar plane with one focus at (0, 0) and a vertex at (-3, -Pi /2)?

2. Hello, kaleigh.thomas!

How do i create an equation for a parabola in a polar plane
with its focus at $(0, 0)$ and its vertex at $\left(\text{-}3,\:\text{-}\frac{\pi}{2}\right)$ ?
You are expected to be familiar with polar equations of conics.

A conic with its focus at (0,0) has the form:

. . $r \;=\;\frac{ke}{1 +e\sin\theta}\:\text{ or }\:\frac{ke}{1 + e\cos\theta}\quad\text{ where: }\;\begin{Bmatrix}e < 1 & \text{ellipse} \\ e = 1 & \text{parabola} \\ e > 1 & \text{hyperbola} \end{Bmatrix}$

From the given information, we know the orientation of the parabola.
Code:
                |
*
*    |    *
*       |       *
- - * - - - - + - - - - * - -
*          |          *
|
*           |           *
|

We know that $e = 1$, and the parabola is "vertical".
Its equaton contains $\sin\theta$.

For the vertex to be at $\left(3,\,\tfrac{\pi}{2}\right)$, we have $k = 6$.

. . Therefore: . $r \;=\;\frac{6}{1+\sin\theta}$