If the tangent at the point $(a t^{2}, a t^{3})$ on the curve $a y^{2} = x^{3}$ meets the curve again at Q.then the co-ordinates of Q is/are

(- \frac{1}{4} a t^{2}, - \frac{1}{8} a t^{3}) .

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(\frac{1}{4} a t^{2}, - \frac{1}{8} a t^{3}) .

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(\frac{1}{4} a t^{3}, - \frac{1}{8} a t^{2}) .

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(- \frac{1}{4} a t^{3}, - \frac{1}{8} a t^{3}) .

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Solution

The correct option is C $(\frac{1}{4} a t^{2}, - \frac{1}{8} a t^{3}) .$
Given, $a y^{2} = x^{3}$
$\Rightarrow 2 a y \frac{d y}{d x} = 3 x^{2}$
$\Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{2 a y}$
Thus slope of tangent at point $(a t^{2}, a t^{3})$ is $m = \frac{3. a^{2} t^{4}}{2 a . a t^{3}} = \frac{3}{2} t$
Therefore equation of tangent at the same point is $(y - a t^{3}) = \frac{3}{2} t (x - a t^{2})$
Now let this line again intersect the given curve at $(a t_{1}^{2}, a t_{1}^{3})$
$\Rightarrow (a t_{1}^{3} - a t^{3}) = \frac{3}{2} t (a t_{1}^{2} - a t^{2})$
$\Rightarrow (t_{1}^{3} - t^{3}) = \frac{3}{2} t (t_{1}^{2} - t^{2})$
$\Rightarrow 2 (t_{1}^{2} + t^{2} - t t_{1}) = 3 t (t_{1} - t) \Rightarrow 2 t_{1}^{2} - t^{2} - t t_{1} = 0$
$\Rightarrow (t - t_{1}) (2 t_{1} + t) = 0 \Rightarrow t_{1} = - \frac{1}{2} t$
Hence required point is, $(\frac{1}{4} a t^{2}, - \frac{1}{8} a t^{3})$

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