Ellipse problem

**arze** · May 7th 2010, 07:54 PM

The tangent at $P(a\cos\theta, b\sin\theta)$ to the ellipse
$b^2x^2+a^2y^2=a^2b^2$
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.

$Q(0,b\csc\theta)$
S'(-ae,0) and S(ae,0)
so gradient of QS' $=\frac{b\csc\theta}{ae}=\frac{a\sin\theta}{b\cos\t heta}$
this means $e=\frac{b^2\cos\theta}{a^2\sin^2\theta}$
that would make ae= $\frac{b^2\cos\theta}{a\sin^2\theta}$
but in order for PS to be vertical, ae needs to be $a\cos\theta$
i.e. e= $\cos\theta$

**sa-ri-ga-ma** · May 7th 2010, 09:58 PM

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

**arze** · May 8th 2010, 02:33 AM

Originally Posted by sa-ri-ga-ma

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

where is N?

**sa-ri-ga-ma** · May 8th 2010, 02:53 AM

Originally Posted by arze

where is N?

Normal at P which cuts the x axis at N.

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