The locus of point of intersection of two normals drawn to the parabola $y^{2} = 4 a x$ which are at right angles is

y^{2} = a (x - 3 a)

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y^{2} = a (x - a)

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y^{2} = 3 a (x - 2 a)

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y^{2} = 2 a (x - 2 a)

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Solution

The correct option is A $y^{2} = a (x - 3 a)$
Let the two points at which normals are drawn be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$
Since, the normals are perpendicular,
$∴ (- t_{1}) (- t_{2}) = - 1$
$\Rightarrow t_{1} t_{2} = - 1$
Now the points of intersection of normals will be $(2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}), - a t_{1} t_{2} (t_{1} + t_{2}))$
Let the locus be $(h, k)$
$\Rightarrow h = 2 a + a (t_{1}^{2} + t_{2}^{2} - 1)$
$\Rightarrow h = a (t_{1}^{2} + t_{2}^{2} + 1)$
$\Rightarrow h = a ((t_{1} + t_{2})^{2} - 2 t_{1} t_{2} + 1)$
$\Rightarrow h = a ((t_{1} + t_{2})^{2} + 3) . . . (1)$
And $k = a (t_{1} + t_{2}) . . . (2)$

From eqn $(1)$ and $(2)$ , we get
$h = a [{(\frac{k}{a})}^{2} + 3]$
$\Rightarrow a (h - 3 a) = k^{2}$
So, locus is $y^{2} = a (x - 3 a)$

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