Consider the parabola $y^{2} = 4 x$ . A chord PQ is normal to the parabola at P $(4, 4)$ , and the angle $P S Q = 90^{o}$ where S is its focus.
Tangents drawn to the parabola $y^{2} = 4 x$ at P and Q, intersect at T. R is the midpoint of TQ then

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$T S R = 90^{o}$

$S Q = 2 S P$

Length of the common chord of circum circles of the triangles SPQ, SQR is 10

$S T = 2 S R$

answer is A, B, C, D.

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Detailed Solution

ABCD
$P (4, 4)$
$Q (9, - 6)$
$R (- 6, - 1)$
$T (\frac{3}{2}, - \frac{7}{2})$
Length of common chord of the circum circles of the triangles $S P Q$ & $R S Q$ is $S Q = 10$

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