P and Q are two distinct points on the parabola, $y^{2} = 4 x,$ with parameters t and $t_{1}$ respectively. If the normal at P passes through Q, then the minimum value of $t_{1}^{2}$ is :

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Detailed Solution

We know that normal at $t_{1}$ meets the parabola at $t_{2}$

then $t_{2} = - t_{1} - \frac{2}{t_{1}}$

By applying this formula we get

$\Rightarrow t_{1} = - t - \frac{2}{t} t^{2} + t_{1} t + 2 = 0 ∆ \geq 0 \Rightarrow {t_{1}}^{2} \geq 8 S o, \min i m u m v a l u e of {t_{1}}^{2} = 8$

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