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

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*  *
*  *
  *     +     *  
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We know that $\displaystyle e = 1$, and the parabola is "vertical".
Its equaton contains $\displaystyle \sin\theta$.
For the vertex to be at $\displaystyle \left(3,\,\tfrac{\pi}{2}\right)$, we have $\displaystyle k = 6$.
. . Therefore: .$\displaystyle r \;=\;\frac{6}{1+\sin\theta} $