A straight line is drawn parallel to the conjugate axis of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to meet it and the conjugate hyperbola in the points P & Q. Show that the tangents at P & Q meet on the curve $\frac{y^4}{b^4}\left(\frac{y^2}{b^2}-\frac{x^2}{a^2}\right)=4\frac{x^2}{a^2}$.
So here is my approach:
taking the point P as
$P\left(a.sec\theta, b.tan\theta \right)$,
since the line was parallel to the conjugate axis, i.e., the y axis, the x coordinate of Q will also be $(a.sec\theta)$ and by putting this in the equation of the conjugate hyperbola, I was able to get the y-coordinate of Q and finally, the point was:
$Q\left(a.sec\theta, b\sqrt{1+sec^2\theta}\right)$
So finally I got the respective tangents by putting the values of P and Q in the general tangent equations:
Tangent at P: $\frac{x.sec\theta}{a}-\frac{y.tan\theta}{b}=1$
Tangent at Q: $\frac{x.sec\theta}{a}-\frac{y.\sqrt{1+sec^2\theta}}{b}=-1$
This is where I got stuck as solving these equations yields a weird result. Can someone please let me know how I can proceed or if this approach is wrong?
Hint:
Let $P(a\tan u,b\sec u);Q(a\tan v,b\sec v)$
The gradient of $PQ$ is $$\dfrac{b(\sec u-\sec v)}{a(\tan v-\tan u)}=\dfrac{b(\cos v-\cos u)}{a\sin(u-v)}=\dfrac{2b\sin\dfrac{u+v}2}{2a\cos\dfrac{u+v}2}$$
As $PQ|| $ the conjugate axis, $\cos\dfrac{u+v}2=0\implies v=(2n+1)\pi-u$ where $n$ is any integer
$\implies Q(-a\tan u,-b\sec u)$
Now form the equation of the tangents at $P,Q$ and solve for $x,y$