## Precalculus (10th Edition)

Use the fact that $d(F,P)=e\cdot d(D,P)$ and that $d(D,P)=p+r\sin\theta$ to derive and have $r=\dfrac{ep}{1-e\sin\theta}$ Refer to the step-bystep part for the actual derivation.
We know that $d(F,P)=e\cdot d(D,P)$ and that $d(D,P)=p+r\sin\theta$. Hence, $r=e\cdot(p+r\sin\theta)\\r=ep+er\sin\theta\\r-er\sin\theta=ep\\r(1-e\sin\theta)=ep\\r=\frac{ep}{1-e\sin\theta}$