1. Ellipse problem

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=$\displaystyle \cos\theta$

2. Normal PN bisects the angle S'PS.
Compare the triangles PNS and QPS'.
They are similar.
So $\displaystyle \angle{PNS} = \angle{QPS'}$
But $\displaystyle \angle{QPS} + \angle{S'PN} = 90 degrees$
Hence $\displaystyle \angle{PNS} + \angle{NPS} = 90 degrees$
So the PSN is a right angled triangle.

3. Originally Posted by sa-ri-ga-ma
Normal PN bisects the angle S'PS.
Compare the triangles PNS and QPS'.
They are similar.
So $\displaystyle \angle{PNS} = \angle{QPS'}$
But $\displaystyle \angle{QPS} + \angle{S'PN} = 90 degrees$
Hence $\displaystyle \angle{PNS} + \angle{NPS} = 90 degrees$
So the PSN is a right angled triangle.
where is N?

4. Originally Posted by arze
where is N?
Normal at P which cuts the x axis at N.