The normal to the curve $x = a t^{2}, y = 2 a t$ at the point $P (t)$ meets the curve again at $Q (t^{'})$ . Then $t^{'}$ is

t + \frac{1}{t}

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- t - \frac{2}{t}

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t + \frac{2}{t}

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t - \frac{1}{t}

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Solution

The correct option is C $- t - \frac{2}{t}$
Normal at $x = a t^{2}, y = 2 a t$ is given by,
$y + t x = 2 a t + a t^{3}$
Given it also passes through $Q (t^{'})$ which is $x = a t^{' 2}, y = 2 a t^{'}$
$\Rightarrow 2 a t^{'} + a t^{' 2} t = 2 a t + a t^{3}$
$\Rightarrow 2 (t^{'} - t) = t (t^{2} - t^{' 2}) = t (t + t^{'}) (t - t^{'})$
$\Rightarrow 2 = - t (t + t^{'})$ (since $t \neq t^{'})$
$\Rightarrow t^{'} = - \frac{2}{t} - t$

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