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 ⇒

<br /> (\frac{b^2}{a^2})(\frac{\sin(\phi)}{1-\cos(\phi)})(\frac{\sin(\theta)}{1-\cos(\theta)})=-1<br />

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

<br /> (a\frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}, 0)<br />

Presumably I should be able to manipulate the first expression to yield a constant value for <br /> \frac{\cos(\frac{\theta-\phi}{2})}{\cos(\frac{\theta+\phi}{2})}<br /> 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

<br /> a(\frac{a^2+b^2}{a^2-b^2},0)<br />

Thanks for your help.

 

[haruspex]

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

<br /> a(\frac{a^2+b^2}{a^2-b^2},0)<br />

Thanks for your help.

Well done.