Math Help - Geometry question

1. Geometry question

Show that an equation of the chord joining the points $P(acos\omega, bsin\omega) and Q(acos\theta, b sin\theta)$ on the ellipse with equation $b^2x^2+a^2y^2=a^2b^2 is Bxcos\frac{1}{2}(\theta+\omega)+aysin\frac{1}{2}(\ theta +\omega)=abcos\frac{1}{2}(\theta-\omega)$

I just can't seem to make it work. Help please!!!

2. Originally Posted by speedfreak
Show that an equation of the chord joining the points $P(acos\omega, bsin\omega) and Q(acos\theta, b sin\theta)$ on the ellipse with equation $b^2x^2+a^2y^2=a^2b^2 is Bxcos\frac{1}{2}(\theta+\omega)+aysin\frac{1}{2}(\ theta +\omega)=abcos\frac{1}{2}(\theta-\omega)$

I just can't seem to make it work. Help please!!!
HI

First of all , find the gradient of PQ

$m=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}$

Then make use of this formula : $y-y_1=m(x-x_1)$

$y-b\sin \omega=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}(x-a\cos \omega)$

$(y-b\sin \omega)(a\cos \omega-a\cos \theta)=(b\sin \omega-b\sin \theta)(x-a\cos \omega)$

$ya\cos \omega-ya\cos \theta-bx\sin \omega+bx\sin \theta=ba\sin \theta\cos \omega-ba\sin \omega\cos \theta$

$ya(\cos \omega-\cos \theta)+bx(\sin \theta-\sin \omega)=ba(\sin \theta\cos \omega-\sin \omega\cos \theta)$

Then use the sin and cos formulas .

$
ya[-2\sin \frac{\omega+\theta}{2}\sin \frac{\omega-\theta}{2}]+bx[2\cos \frac{\theta+\omega}{2}\sin \frac{\theta-\omega}{2}]=ba\sin (\theta-\omega)
$

$ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\sin (\theta-\omega)}{-2\sin \frac{1}{2}(\theta-\omega)}$

$
ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\cdot 2\sin \frac{\omega-\theta}{2}\cos \frac{\omega-\theta}{2}}{2\sin \frac{\omega-\theta}{2}}
$

Then cancel the like terms to get the equation . Phew ..

3. Originally Posted by mathaddict
HI

First of all , find the gradient of PQ

$m=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}$

Then make use of this formula : $y-y_1=m(x-x_1)$

$y-b\sin \omega=\frac{b\sin \omega-b\sin \theta}{a\cos \omega-a\cos \theta}(x-a\cos \omega)$

$(y-b\sin \omega)(a\cos \omega-a\cos \theta)=(b\sin \omega-b\sin \theta)(x-a\cos \omega)$

$ya\cos \omega-ya\cos \theta-bx\sin \omega+bx\sin \theta=ba\sin \theta\cos \omega-ba\sin \omega\cos \theta$

$ya(\cos \omega-\cos \theta)+bx(\sin \theta-\sin \omega)=ba(\sin \theta\cos \omega-\sin \omega\cos \theta)$

Then use the sin and cos formulas .

$
ya[-2\sin \frac{\omega+\theta}{2}\sin \frac{\omega-\theta}{2}]+bx[2\cos \frac{\theta+\omega}{2}\sin \frac{\theta-\omega}{2}]=ba\sin (\theta-\omega)
$

$ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\sin (\theta-\omega)}{-2\sin \frac{1}{2}(\theta-\omega)}$

$
ay\sin \frac{\omega+\theta}{2}+bx\cos \frac{\omega+\theta}{2}=\frac{ba\cdot 2\sin \frac{\omega-\theta}{2}\cos \frac{\omega-\theta}{2}}{2\sin \frac{\omega-\theta}{2}}
$

Then cancel the like terms to get the equation . Phew ..
Thanks most helpful!!
I am also stuck on the second part of the question:
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.
So far I have work out the gradients of both AP and AQ and deduced that because they are perpendicular:
$b^2\sin\theta\sin\omega = a^2(\cos\theta + \cos\omega - 1 - \cos\theta\cos\omega)$
and also that $xcos\frac{\theta+\omega}{2} = acos\frac{\theta-\omega}{2}$