1. ## Ellipse problem

Hi,

So let $P(a\cos\theta, b\sin\theta)$ and $Q(a\cos\phi, b\sin\phi)$ be points on the standard ellipse. I'm trying to show that if $PQ$ subtends a right angle at the point $(a, 0)$ then the chord $PQ$ passes through a fixed point on the x-axis.

So far I've gotten that $x = \frac{a\sin(\theta + \phi)}{\sin\phi - \sin\theta}$ and I need to show that this is invariant. The condition (that P - (a, 0) - Q forms a right angle) gives me

$b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)$

I've messed around with this for well over 2 hours and all I've gotten into is a terrible mess with some very tedious algebra and nothing to show for it!

Any help would be much appreciated, thanks

Stonehambey

2. Originally Posted by Stonehambey
Hi,

So let $P(a\cos\theta, b\sin\theta)$ and $Q(a\cos\phi, b\sin\phi)$ be points on the standard ellipse. I'm trying to show that if $PQ$ subtends a right angle at the point $(a, 0)$ then the chord $PQ$ passes through a fixed point on the x-axis.

So far I've gotten that $x = \frac{a\sin(\theta + \phi)}{\sin\phi - \sin\theta}$ and I need to show that this is invariant. The condition (that P - (a, 0) - Q forms a right angle) gives me

$b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)$

I've messed around with this for well over 2 hours and all I've gotten into is a terrible mess with some very tedious algebra and nothing to show for it!

Any help would be much appreciated, thanks

Stonehambey
when i calculated for x i got $x = \frac{a\sin(\theta - \phi)}{\sin\phi - \sin\theta}$. re check yours
now all it needs to be proved is that this is a constant when $b^2 \sin\theta \sin \phi = -a^2 (\cos\theta - 1)(\cos\phi - 1)$
the last equation when manipulated becomes $b^2 + a^2(\tan \theta /2)(\tan \phi /2)=0$
this means $\phi$ is a function of $\theta$.
differentiate this both sides and you will get $\frac{d \phi}{d \theta}= \frac{- \sin \phi}{\sin \theta}$
simplyfy x by using difference formulas, that is $x= \frac{a\sin(\theta - \phi)}{\sin\phi - \sin\theta}=a\frac{\cos {(\theta - \phi)/2}}{\cos{(\theta + \phi )/2}}$
now x is completely a function of $\theta$. compute $\frac{dx}{d \theta}$ buy using the value of $\frac{d \phi}{d \theta}$ calculated above.
you will get $\frac{dx}{d \theta}=0$ which will prove it.

I managed to get right to the end of that, but when I got to finding $\frac{dx}{d\theta}$ I got

$\frac{dx}{d\theta} = \left(\cos\left( \frac{\theta + \phi}{2} \right)\left[-\sin\left( \frac{\theta - \phi}{2}\right)\left( \frac{1}{2} + \frac{\sin\phi}{2\sin\theta} \right)\right] - \cos\left( \frac{\theta - \phi}{2} \right)\left[-\sin\left( \frac{\theta + \phi}{2}\right)\left( \frac{1}{2} - \frac{\sin\phi}{2\sin\theta} \right)\right]\right) / \cos^2\left( \frac{\theta + \phi}{2} \right)$

which looks horrible! How do I show that this is zero? Or have I made a mistake somewhere?

Thanks

4. Originally Posted by Stonehambey

I managed to get right to the end of that, but when I got to finding $\frac{dx}{d\theta}$ I got

$\frac{dx}{d\theta} = \left(\cos\left( \frac{\theta + \phi}{2} \right)\left[-\sin\left( \frac{\theta - \phi}{2}\right)\left( \frac{1}{2} + \frac{\sin\phi}{2\sin\theta} \right)\right] - \cos\left( \frac{\theta - \phi}{2} \right)\left[-\sin\left( \frac{\theta + \phi}{2}\right)\left( \frac{1}{2} - \frac{\sin\phi}{2\sin\theta} \right)\right]\right) / \cos^2\left( \frac{\theta + \phi}{2} \right)$

which looks horrible! How do I show that this is zero? Or have I made a mistake somewhere?

Thanks
I too was getting something like this (although i have not checked this term by term but still .... )
now to prove that this is zero use $\sin \theta + \sin \phi = 2 \sin[ (\theta+\phi)/2] \cos[ (\theta-\phi)/2]$
and
$\sin \theta -\sin \phi=2 \sin [(\theta-\phi)/2] \cos [(\theta +\phi)/2]$
this does it.