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Chord of Contact

A variable ch...

Question

A variable chord $P Q$ of the parabola $y^{2} = 4 a x$ is drawn parallel to line $y = x$ . Then the locus of point of intersection of normals at $P$ and $Q$ is:

A

2 x - y - 12 a = 0

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B

2 x + y - a = 0

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C

x - 2 y - 4 a = 0

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D

3 x - y - 8 a = 0

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Solution

The correct option is A $2 x - y - 12 a = 0$
Equation of normal's at $P (a t_{1}^{2}, 2 a t_{2})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ are
$y + x t_{1} = 2 a t_{1} + a t_{1}^{3}$ and
$y + x t_{2} = 2 a t_{2} + a t_{2}^{3}$
Point of intersection of both lines is
$x = 2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2})$ and
$y = - a t_{1} t_{2} (t_{1} + t_{2})$
$∵$ Slope of chord
$P Q = \frac{2 a t_{1} - 2 a t_{2}}{a t_{1}^{2} - a t_{2}^{2}} = 1$
$∴ t_{1} + t_{2} = 2$
$\Rightarrow y = - a t_{1} t_{2} (t_{1} + t_{2}) \Rightarrow t_{1} t_{2} = \frac{- y}{2 a}$
$x = 2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}) \Rightarrow x = 2 a + a ((t_{1} + t_{2})^{2} - t_{1} t_{2})$
$\Rightarrow x = 2 a + a (4 + \frac{y}{2 a})$
$2 x - y - 12 a = 0$

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