The question is:
$PQ$ is a chord joining the points $\phi_1$ and $\phi_2$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $\phi_1\,+\,\phi_2 = 2\alpha$, where $\alpha$ is constant, prove that $PQ$ touches the hyperbola $\frac{x^2}{a^2}\cos^2\alpha-\frac{y^2}{b^2}=1$.
I found out the equation of the chord,
$$\frac{x(\tan\phi_1-\tan\phi_2)}{a}-\frac{y(\sec\phi_1-\sec\phi_2)}{b}+(\sec\phi_1\tan\phi_2-\sec\phi_2\tan\phi_1)=0$$
But how can I show that this line touches the required hyperbola?
Hint: Solve the equation of the chord in terms of $\frac{y}{b}$. Then calculate $\frac{y^2}{b^2}$ and using trigonometric identities and the fact that $\phi_1+\phi_2=2\alpha$ you should obtain that $\frac{y^2}{b^2}=\frac{x^2}{a^2}cos^2{\alpha}-1$. This proves that the system of the two equations has a solution, namely the chord touches the hyperbola $\frac{x^2}{a^2}cos^2{\alpha}-\frac{y^2}{b^2}=1$. I guess by $\phi_i$ being a point you mean that $x=asec(\phi_i)$, $y=btan(\phi_i)$ is the point for $i=1,2$