The normal chord at apoint t on the parabola $y^{2} = 4 a x$ subtends a right at the vertex. Then $t^{2}$ is equal to

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\sqrt{2}

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None of these

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Solution

The correct option is A 2
Let the normal at $P$ meets the parabola again at $Q$ then
$t^{'} = - t - \frac{2}{t}$
$t t^{'} = - t^{2} - 2$

Slope of $O P = \frac{2 a t - 0}{a t^{2} - 0} = \frac{2}{t}$

Similarly , Slope of $O Q = \frac{2}{t^{'}}$

$O P i s p e r p e n d i c u l a r t o O Q$

$\frac{2}{t} \times \frac{2}{t^{'}} = - 1$

$t t^{'} = - 4$

$- 4 = - t^{2} - 2$

$t^{2} = 2$

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