Ellipse chord subtending a right angle

[Appleton]

Homework Statement

Show that the equation of the chord joining the points $P(a\cos(\phi), b\sin(\phi))$and $Q(a\cos(\theta), b\sin(\theta))$ on the ellipse $b^2x^2+a^2y^2=a^2b^2$ is $bx cos\frac{1}{2}(\theta+\phi)+ay\sin\frac{1}{2}(\theta+\phi)=ab\cos\frac{1}{2}(\theta-\phi)$.
Prove that , if the chord PQ subtends a right angle at the point (a,0), then PQ passes through a fixed point on the x axis.

Homework Equations

The Attempt at a Solution

The second part of the question is where I am having difficulty.

The question seems to suggest a parametric as opposed to cartesian approach.

PQ subtends a right angle ⇒

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

The coordinates of the intersection of PQ with the x-axis are

$(a\frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}, 0)$

Presumably I should be able to manipulate the first expression to yield a constant value for $\frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}$ however, I don't seem to be making much progress in this respect.

 

[haruspex]

Since one of the expressions involves half angles, try substituting theta=2A, phi=2B everywhere.

 

[Appleton]

That seems to have done the trick. So the point of intersection is the constant value

$a(\frac{a^2+b^2}{a^2-b^2},0)$

Thanks for your help.

 

[haruspex]

That seems to have done the trick. So the point of intersection is the constant value

$a(\frac{a^2+b^2}{a^2-b^2},0)$

Thanks for your help.

Well done.