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Q.

PQ and PR are tangents to parabola $y^{2} = 8 x$ where P lies on x-axis and Q(2,4) and R are points on the parabola. If normal to parabola at R intersects the parabola again at S, then point of intersection of tangents at Q and S is

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a

(8, 6)

b

(-6, 4)

c

(6, 8)

d

(3, 4)

answer is B.

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$\begin{array}{l} y^{2} = 8 x, Q (2, 4) a = 2 \\ P ({at}_{1} t_{2}, a (t_{1} + t_{2})) lies on x - axis \\ \Rightarrow t_{1} + t_{2} = 0 \\ \Rightarrow parameter of Q is t = 1 \\ \Rightarrow parameter of R is t = - 1 \\ \Rightarrow parameter of S is t_{2} = - t_{1} - \frac{2}{t_{1}} = 1 + 2 = 3 \\ Point of intersection of tangents at Q (1), S (3) \\ = (a t_{1} t_{2}, a (t_{1} + t_{2})) = (2 \times 3, 2 (1 + 3)) = (6, 8) \end{array}$

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