There's a very self-contained statement in the screenshot below from a book on conic sections. Note that KP and XS are orthogonal to KF. Think of S and P as a arbitrary points on their respective straight lines and P' is constructed as described. For the life of me I'm not seeing why SP'/SP = KP'/KP.
From the comments:
This proof uses the external angle bisector theorem. We apply the theorem to the triangle $\triangle PSP'$, where $KS$ is the external bisector at vertex $S$. The theorem implies that
$$ \frac{\lvert SP'\rvert}{\lvert SP\rvert} = \frac{\lvert KP'\rvert}{\lvert KP\rvert}. $$