The equation of the locus of the point of intersection of two normals to the parabola $y^{2} = 4 a x$ which are perpendicular to each other is

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Solution

The correct option is A
Let $P (x_{1}, y_{1})$ be the point of intersectin of the two perpendicular normals at $A (t_{1}), B (t_{2})$ on the parabola $y^{2} = 4 a x$
let $t_{3}$ be the foot of the third normal through P.Equation of a normal at t to the parabola is $y + x t = 2 a t + a t^{3}$
If this normal passes through P then $y_{1} + x_{1} t = 2 a t + a t^{3} \Rightarrow a t^{3} + (2 a - x_{1}) t - y_{1} = 0 \to (1)$
Now $t_{1}, t_{2}, t_{3}$ are the roots of (1). $∴ t_{1} t_{2} t_{3} = \frac{y_{1}}{a}$
Slope of the normal at $t_{1}$ is $- t_{1}$ , slope of normal at $t_{2}$ is $- t_{2}$
Normals at $t_{1}$ and $t_{2}$ are perpendicular $\Rightarrow (- t_{1}) (- t_{2}) = - 1 \Rightarrow t_{1} t_{2} t_{3} = - t_{3} \Rightarrow \frac{y_{1}}{a} = - t_{3} \Rightarrow t_{3} = \frac{y_{1}}{a}$
$t_{3}$ is a root of (1) $\Rightarrow a {(- \frac{y_{1}}{a})}^{3} + (2 a - x_{1}) (- \frac{y_{1}}{a}) - y_{1} = 0 \Rightarrow - \frac{y_{1}^{3}}{a^{2}} - \frac{(2 a - x_{1}) y_{1}}{a} - y_{1} = 0$
$\Rightarrow y_{1}^{2} + a (2 a - x_{1}) + a^{2} = 0 \Rightarrow y_{1}^{2} = a (x_{1} - 3 a) ∴$ The locus of P is $y^{2} = a (x - 3 a)$

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