Parabola equation from cartesian to polar representation

Question

I've got the following equation:
0) $ \frac{(y-y_p)^{2}}{4\cdot(x-x_p)} = p $

I'd like now to convert this expression to a polar representation. For this I got back to the basic rules:
1) $x = r\cdot cos\theta$
2) $y = r\cdot sin \theta$
3) $r = \sqrt{x^{2}+y{2} }$
4) $\theta = tan^{-1}(\frac{y}{x})$

Applying rule 1) and 2) to y and x only, we get:
$\frac{(r\cdot sin \theta-y_p)^{2}}{4\cdot(r\cdot cos\theta-x_p)} = p$

Now I got stuck with the following question: how are the parameters $x_p$, $y_p$ and $p$ treated?

However I'm thinking that my approach is wrong or that at least something is missing in order to fully convert equation 0) to polar coordinates.

Thank you in advance for any hints and with best regards Dan

Answer 1

extracting variables $x,y$, we obtain $$\frac{(y-y_p)^2}{4(x-x_p)}=p$$ The trick consists on performing the polar transformation on the point $(x_p,y_p)$. Then $y-y_p=r\sin\theta$ and $x-x_p=r\cos\theta$. Finding $$r\tan(\theta)\sin(\theta)=4p$$

Answer 2

HINT $$ \frac{(y-y_p)^{2}}{4\cdot(x-x_p)} = p $$ and substituting $y-y_p=r\sin\theta$ and $x-x_p=\cos\theta$, you'll find $$ r(\theta)=4p\frac{\cos\theta}{\sin^2\theta} $$