The tangent at $\displaystyle P(a\cos\theta, b\sin\theta)$ to the ellipse

$\displaystyle b^2x^2+a^2y^2=a^2b^2$

cuts the y-axis at Q. The normal at P is parallel to the line joining Q to one focus S'. if S is the other focus, show that PS is parallel to the y-axis.

$\displaystyle Q(0,b\csc\theta)$

S'(-ae,0) and S(ae,0)

so gradient of QS' $\displaystyle =\frac{b\csc\theta}{ae}=\frac{a\sin\theta}{b\cos\t heta}$

this means $\displaystyle e=\frac{b^2\cos\theta}{a^2\sin^2\theta}$

that would make ae=$\displaystyle \frac{b^2\cos\theta}{a\sin^2\theta}$

but in order for PS to be vertical, ae needs to be $\displaystyle a\cos\theta$

i.e. e=$\displaystyle \cos\theta$