A variable parabola $y^{2} = 4 a x$ , ‘a’ being the parameter $(a \neq \frac{- 1}{4})$ meets the curve $y^{2} + x - y - 2 = 0$ at two points .The locus of the point of intersection of tangents at these points is

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$x - 4 y - 1 = 0$

$2 x - y + 1 = 0$

$x - 4 y + 2 = 0$

$x - 2 y - 4 = 0$

answer is B.

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Detailed Solution

Point of intersection $(a t^{2}, 2 a t)$ lies on $y^{2} + x - y - 2 = 0$
$4 a^{2} t^{2} + a t^{2} - 2 a t - 2 = 0 \Rightarrow (4 a^{2} + a) t^{2} - 2 a t - 2 = 0$
$t_{1} + t_{2} = \frac{2 a}{4 a^{2} + a}$
$t_{1} t_{2} = \frac{- 2}{(4 a^{2} + a)}$
Point of intersection of tangents, (x, y) = $(a t_{1} t_{2}, a (t_{1} + t_{2}))$
$x = \frac{- 2 a}{4 a^{2} + a}; y = \frac{2 a^{2}}{4 a^{2} + a} \to \frac{y}{x} = - a ∴ x = \frac{- 2 (- y / x)}{4 {(\frac{y}{x})}^{2} + (\frac{- y}{x})} \Rightarrow x = \frac{2 x y}{4 y^{2} - x y} \Rightarrow 1 = \frac{2}{4 y - x} \Rightarrow 4 y - x = 2$