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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
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The correct answers are:
$t=-\operatorname{cosec} \theta$, $t=2 \tan \theta$
$\left(a,\right.$Equation of tangent to $y^{2}=4 a x$ at $\left(a t^{2}, 2 a t\right)$ will be
$$
x-y t=-a t^{2}
$$
Also, equation of normal to $x^{2}-y^{2}=a^{2}$ at $(a \sec \theta, a \tan \theta)$ will be
$\frac{x}{a \sec \theta}+\frac{y}{a \tan \theta}=2$
$\Rightarrow x+y \operatorname{cosec} \theta=2 a \sec \theta$
since, Eqs. (i) and (ii) are identical.
$\therefore t=-\operatorname{cosec} \theta$ or $t=2 \tan \theta$
$$
x-y t=-a t^{2}
$$
Also, equation of normal to $x^{2}-y^{2}=a^{2}$ at $(a \sec \theta, a \tan \theta)$ will be
$\frac{x}{a \sec \theta}+\frac{y}{a \tan \theta}=2$
$\Rightarrow x+y \operatorname{cosec} \theta=2 a \sec \theta$
since, Eqs. (i) and (ii) are identical.
$\therefore t=-\operatorname{cosec} \theta$ or $t=2 \tan \theta$
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