$T P$ & $T Q$ are tangents to the parabola and the normals at $P$ & $Q$ meet at a point $R$ on the curve. The centre of the circle circumscribing the $△ T P Q$ lies on a parabola. Find equation of parabola.

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

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

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

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

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Solution

The correct option is C $2 y^{2} = a (x - a)$
Let parabola $y^{2} = 4 a x$
Let the points on parabola be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$
Point of intersection of tangent at $P$ & $Q$ is $T$ $(a t_{1} t_{2}, a (t_{1} + t_{2}))$
Normal at $P$ & $Q$ meet again in the parabola so relation between $t_{1}$ and $t_{2}$ is
$- t_{1} - \frac{2}{t_{1}} = - t_{2} - \frac{2}{t_{2}}$
$t_{1} t_{2} = 2$
Equation of line perpendicular to $T P$ & passing through mid point of $T P$ is
$2 y - a (3 t_{1} + t_{2}) = - t_{1} (2 x - a (2 + t_{1}^{2}))$ .... (1)
$2 y + 2 x t_{1} = a (3 t_{1} + t_{2}) + a t_{1} (2 + t_{1}^{2})$
Similarly equation of line passing through mid point of $T Q$ and $⊥$ to $T Q$ is
$2 y + 2 x t_{2} = a (3 t_{2} + t_{1}) + a t_{2} (2 + t_{2}^{2})$ .... (2)
From (1) & (2) using $t_{1} t_{2}$ = 2
Eliminating $t_{1}$ and $t_{2}$ we get the locus of circumcentre
$2 y^{2} = a (x - a)$

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