Thank you for reporting, we will resolve it shortly
Q.
If the tangent to $y^{2}=4 a x$ at the point $\left(a t^{2}, 2 a t\right)$ where $|t|>1$ is a normal to $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta)$, then
Equation of tangent to $y^{2}=4 a x$ at $\left(a t^{2}, 2 a t\right)$ will be
$x-y t=-a t^{2}\,\,\,\,...(i)$
Also, equation of normal to $x^{2}-y^{2}=a^{2}$ at
$(a \sec \theta, a \tan \theta)$ wiil be
$\frac{x}{a \sec \theta}+\frac{y}{a \tan \theta}=2 $
$\Rightarrow x+y\, \text{cosec} \theta=2 a \sec \theta\,\,\,\,...(ii)$
Since, Eqs. (i) and (ii) are identical.
$\therefore t=-\text{cosec} \,\theta$ or $t=2 \tan \theta$