Find the locus of the point of intersection of normals drawn at the extremities of a focal chord of the parabola $y^{2} = 4 a x .$

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

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

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

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y^{2} = \frac{a}{6} (x - 3 a)

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Solution

The correct option is A $y^{2} = a (x - 3 a)$
Point of intersection of the normals drawn at the ends of a chord of $y^{2} = 4 a x$ is $(x, y) = [a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2), - a t_{1} t_{2} (t_{1} + t_{2})]$
Since, for focal $t_{1} t_{2} = - 1$
$x = a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2 = a [{(t_{1} + t_{2})}^{2} - t_{1} t_{2} + 2] = a [{(t_{1} + t_{2})}^{2} + 3]$
and $y = - a t_{1} t_{2} (t_{1} + t_{2}) = a (t_{1} + t_{2})$
eliminating $t_{1} + t_{2}$ from $x$ and $y$ , we get
$y^{2} = a (x - 3 a)$

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