If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then If a  chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \beta=\tan\alpha(4\sec^2\alpha-1)$.

The chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$  is given by $x\cos\frac{\alpha-\beta}{2}-y\sin\frac{\alpha+\beta}{2}=a\cos\frac{\alpha+\beta}{2}$....................$(1)$
And the normal to the hyperbola at $P(a\sec\alpha,a\tan\alpha)$ is given by
$\frac{x}{\sec\alpha}+\frac{y}{\tan\alpha}=2a$.............$(2)$
equation $(1)$ and $(2)$ are the same lines
Comparing them,gives
$\frac{\cos\frac{\alpha-\beta}{2}}{\frac{1}{\sec\alpha}}=\frac{-\sin\frac{\alpha+\beta}{2}}{\frac{1}{\tan\alpha}}=\frac{a\cos\frac{\alpha+\beta}{2}}{2a}$
$\sec\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}}$ and $\tan\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{-2\sin\frac{\alpha+\beta}{2}}$
I am stuck here.I do not know how to proceed further.

As $(2)$ passes through $Q(a\sec\beta,a\tan\beta),$
$$\dfrac{\sec\beta}{\sec\alpha}=2-\dfrac{\tan\beta}{\tan\alpha}$$
Squaring and writing $\tan\alpha=p,\tan\beta=q,$
$$\dfrac{1+q^2}{1+p^2}=4+\dfrac{q^2}{p^2}-\dfrac{4q}p$$
$$\iff4-\dfrac{4q}p=\dfrac{1+q^2}{1+p^2}-\dfrac{q^2}{p^2}$$
$$\iff\dfrac{4(p-q)}p=\dfrac{p^2-q^2}{(1+p^2)p^2}$$
If $p=q$ i.e., $\tan\alpha=\tan\beta,(2)\implies\sec\beta=\sec\alpha\implies P=Q$
So, cancelling $p-q\ne0$ and assuming $p\ne0,$
$$\iff4=\dfrac{p+q}{(1+p^2)p}\iff q=4p^3+3p=p\{4(p^2+1)-1\}$$
