P is a point on the parabola $y^{2} = 4 a x$ whose ordinate equals its abscissa. A normal is drawn to the parabola at P to meet it again at Q. If S is the focus of the parabola. Slope of SP is $m_{1}$ and slope of SQ is $m_{2}$

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$P (4 a, 4 a)$

$Q (9 a, - 6 a)$

$|S P . S Q| = 50 a^{2}$

$m_{1} m_{2} = - 1$

answer is A, B, C, D.

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Detailed Solution

$P (a t_{1}^{2}, 2 a t_{1})$
$\begin{array}{l} g i v e n a t_{1}^{2} = 2 a t_{1} \\ t_{1} = 2 \\ P = (4 a, 4 a) \\ t_{2} = - t_{1} - \frac{2}{t_{1}} = - 2 - \frac{2}{2} = - 2 - 1 = - 3 \end{array}$
$Q (a t_{2}^{2}, 2 a t_{2}) = (a (a), 2 a (- 3)) = (9 a, - 6 a)$
$\begin{array}{l} P (4 a, 4 a) Q (9 a, - 6 a) S (a, 0) \\ S P = \sqrt{9 a^{2} + 16 a^{2}} = \sqrt{25 a^{2}} = 5 a \\ S Q = \sqrt{64 a^{2} + 36 a^{2}} = \sqrt{100 a^{2}} = 10 a \\ |S P . S Q| = |5 a .10 a| = 50 a^{2} \\ m_{1} = s l o p e o f S P = \frac{4 a - 0}{4 a - a} = \frac{4 a}{3 a} = \frac{4}{3} \\ m_{2} = s l o p e o f S Q = \frac{- 6 a - 0}{9 a - a} = \frac{- 6 a}{8 a} = \frac{- 3}{4} \\ m_{1} m_{2} = \frac{4}{3} \times \frac{- 3}{4} = - 1 \end{array}$