A normal drawn to parabola $y^{2} = 4 a x$ meet the curve again at Q such that angle subtended by PQ at vertex is $90 °$ , and then coordinates of P can be.

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$(2 a, - 2 \sqrt{2} a)$

$(8 a, 4 a)$

$(8 a, 4 \sqrt{2} a)$

$(2 a, 2 \sqrt{2} a)$

answer is C, D.

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Detailed Solution

Let the normal drawn point P $(a t_{1}^{2}, 2 a t_{1})$ on the parabola $y^{2} = 4 a x$ then normal will intersect the parabola at $Q (a t_{1}^{2}, 2 a t_{2}) and t_{2} - t_{1} - \frac{2}{t_{1}}$ Also, $\frac{2 a t_{1}}{a t_{1}^{2}} \times \frac{2 a t_{2}}{a t_{2}^{2}} = - 1$ $\Rightarrow t_{1} t_{2} = - 4 ∴ \frac{- 4}{t_{1}} = - t_{1} - \frac{2}{t_{1}}$
$\Rightarrow t_{1}^{2} + 2 = 4 and t_{1} = \pm \sqrt{2}$ So point can be $(2 a, \pm 2 \sqrt{2 a})$