The locus of the point of intersection of tangents to $y^{2} = 4 a x$ which intercept a constant length d on the directrix is

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Solution

The correct option is A
Let the tanents at A $(a t_{1}^{2}, 2 a t_{1})$ B $(a t_{1}^{2}, 2 a t_{2})$ on the parabola intersect x + a = 0 at P, Q respectively.
Equation of the tangent at A is $t_{1} y = x + a t_{1}^{2} ∴ P (- a, \frac{a (a t_{1}^{2} - 1)}{t_{1}})$
Similarly Q $(- a, \frac{a (a t_{2}^{2} - 1)}{t_{2}})$
Let R $(x_{1}, y_{1})$ be a point in the locus.
$∴ (x_{1}, y_{1}) = [a t_{1} t_{2}, a (t_{1} + t_{2})] \Rightarrow x_{1} = a t_{1} t_{2}, y_{1} = a (t_{1} + t_{2}) \Rightarrow t_{1} t_{2} = \frac{x_{1}}{a}, t_{1} + t_{2} = \frac{y_{1}}{a}$
$d^{2} = P Q^{2} = {[\frac{a (a t_{1}^{2} - 1)}{t_{1}} - \frac{a (a t_{2}^{2} - 1)}{t_{2}}]}^{2} = {\frac{a^{2} [t_{1}^{2} t_{2} - t_{1} t_{2}^{2} + t_{1}]}{t_{2}^{2} t_{2}^{2}}}^{2} = \frac{a^{2} [(t_{1} - t_{2}) + t_{1} t_{2} (t_{1} - t_{2})]^{2}}{t_{1}^{2} t_{2}^{2}}$
$\Rightarrow d^{2} . d_{1}^{2} t_{2}^{2} = a^{2} [(t_{1} - t_{2}) (1 + t_{1} t_{2})]^{2} = a^{2} [(t_{1} + t_{2})^{2} - 4 t_{1} t_{2}] (1 + t_{1} t_{2})^{2}$
$d^{2} {(\frac{x_{1}}{a})}^{2} = a^{2} [{(\frac{x_{1}}{a})}^{2} - 4 (\frac{x_{1}}{a})] {[1 + \frac{x_{1}}{a}]}^{2} = a^{2} (\frac{y_{1}^{2} - 4 a x_{1}}{a^{2}}) \frac{(a + x_{1})}{a^{2}} \Rightarrow d^{2} x_{1}^{2} (y_{1}^{2} - 4 a x_{1}) (x_{1} + a)^{2}$
$L o c u s o f (x_{1}, y_{1}) i s (y^{2} - 4 a x) (x + a)^{2} = d^{2} x^{2}$

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