The equation of the normal to the curve $(x = a t^{2}), (y = 2 a t)$ at ' $t$ ' point is -

t y = x + a t^{2}

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y + t x - 2 a t - a t^{3} = 0

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

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None of these

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Solution

The correct option is B $y + t x - 2 a t - a t^{3} = 0$
Given $x = a t^{2}, y = 2 a t$

$\Rightarrow \frac{d x}{d t} = 2 a t, \frac{d y}{d t} = 2 a$
Thus slope of normal at point 't' is $= - \frac{d x}{d y} = \frac{d x / d t}{d y / d t} = - t$

Therefore, the equation of normal at $^{'} t^{'}$ is given by,

$(y - 2 a t) = - t (x - a t^{2})$

$\Rightarrow y + t x - 2 a t - a t^{3} = 0$

Hence, option 'B' is correct.

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