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Q.

A quadrilateral ABCD in inscribed in a parabola having equation $y^{2} = 16 x$ such that three of its sides AB, BC, CD passes through the fixed points $(- 2, 0), (3, 0)$ & $(- 6, 0)$ respectively. The fourth side DA passes through a fixed point $(h, 0)$ , then $h =$ ____

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answer is 4.

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Detailed Solution

Equation AB is $y (t_{1} + t_{2}) = x + t_{1} t_{2}$

$y = 0 \Rightarrow x = - \frac{t_{1} t_{2}}{a} \Rightarrow t_{1} t_{2} = \frac{2}{4} t_{2} t_{3} = - \frac{3}{4} t_{3} t_{4} = \frac{6}{4} t_{2} t_{3} t_{4} = \frac{6}{4} t_{2} - \frac{3}{4} t_{4} = \frac{6}{4} t_{2} t_{4} = - 2 t_{2} t_{1} t_{4} = - 2 t_{2} t_{1} t_{1} t_{4} = - 2 (\frac{2}{4}) t_{1} t_{4} = 1 h = a t_{1} t_{4} = 4$

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