# Locus of the midpoint of Hyperbola

I need help with this question;

The points P $(a\sec\theta, b\tan\theta )$ and Q $(a\sec \phi , b\tan\phi )$ lie on the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Show that the locus of the midpoint PQ is $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{y}{b}$$ given that $\theta\ + \phi\ = \frac{\pi}{2}$

I've used midpoint formula to have $x=\frac{a(\cos\theta\ + \cos\phi)}{2(\cos\theta\cos\phi)}$ and $y=\frac{b}{2(\cos\theta\cos\phi)}$

Then solved simultaneously to get an answer nowhere near the solution. Any help is much appreciated. Thanks in advance!

$$\implies Q(a\csc\theta,b\cot\theta)$$
If $R(h,k)$ is the midpoint,
$$\dfrac{2h}a=\sec\theta+\csc\theta=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}$$
$$\dfrac{2k}b=\tan\theta+\cot\theta=\dfrac1{\cos\theta\sin\theta}\iff\cos\theta\sin\theta=?$$
On division, $$\cos\theta+\sin\theta=\dfrac{\dfrac{2h}a}{\dfrac{2k}b}=\dfrac{bh}{ak}$$
Now use $(\cos\theta+\sin\theta)^2=1+2\cos\theta\sin\theta$