Find the equation of normal to the parabola $y^{2} = 4 a x a t (a t^{2}, 2 a t)$ in terms of t, a.

$y = t x + a t^{3} + 2 a t$

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$y = - t x + a t^{3} - 2 a t$

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$y = - t x + a t^{3} + 2 a t$

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$y = - t x + a t^{3} + a t$

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Solution

The correct option is C
$y = - t x + a t^{3} + 2 a t$

we know condition for y=mx+c to be the tangent to $y^{2} = 4 a x i s c = \frac{a}{m}$

So, $y = m x + \frac{a}{m}$ passes through $(a t^{2}, 2 a t)$

$\Rightarrow 2 a t = m a t^{2} + \frac{a}{m}$

$\Rightarrow m^{2} t^{2} - 2 m t + 1 = 0 \Rightarrow m = \frac{1}{t}$

$∴ s l o p e o f t a n g e n t = \frac{1}{t}$

$\Rightarrow s l o p e o f n o r m a l a t (a t^{2}, 2 a t) = - 1_{(\frac{1}{t})} = - t$

$\Rightarrow E q u a t i o n o f n o r m a l w i l l b e y = - t x + c^{'}$

We know, $y = - t x + c^{1} p a s s e s t h r o u g h (a t^{2}, 2 a t)$

$\Rightarrow 2 a t = - t (a t_{2} + c_{1})$

$\Rightarrow c = a t^{3} + 2 a t$

$∴$ Equation of normal to $y^{2} = 4 a x a t (a t^{2}, 2 a t) i s y = - t x + a t^{3} + 2 a t$

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