The point P on the parabola $y^{2} = 4 a x$ for which |PR PQ| is maximum, where R $(- a, 0)$ , Q $(0, a)$ , is

(a, 2 a)

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(a, - 2 a)

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(4 a, 4 a)

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(4 a, - 4 a)

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Solution

The correct option is A $(a, 2 a)$
The point on parabola $y^{2} = 4 a x$ for which $(P R . P Q)$ is maximum where $R (- a, 0), Q (0, a)$ is :
$(0, a)$ lie outside parabola
$(P R - P Q)$ is maximum is PQR is at line
Slope of $P Q =$ slope of $P R$
Let
$P = (a t^{2}, 2 a t) \frac{2 a t - a}{a t^{2}} = \frac{2 a t}{a t^{2} + a} \frac{2 t - 1}{t^{2}} = \frac{2 t}{t^{2} + 1} (t - 1)^{2} = 0 t = 1 P = (a, 2 a)$

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