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Parametric Equation of Normal

A variable ch...

Question

A variable chord $P Q$ of the parabola $y^{2} = 4 a x$ is drawn parallel to $y = x$ . 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 + 10 a = 0

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C

2 x - y - 8 a = 0

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D

2 x - y + 6 a = 0

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Solution

The correct option is A $2 x - y - 12 a = 0$
Eq.s of chords at $P (a {t_{1}}^{2}, 2 a t_{1}), Q ((a {t_{2}}^{2}, 2 a t_{2})$ are given by
$y + x t_{1} = 2 a t_{1} + a {t_{1}}^{3}$ ...... $(1)$
$y + x t_{2} = 2 a t_{2} + a {t_{2}}^{3}$ ......... $(2)$
Slope of PQ $= \frac{2 a t_{1} - 2 a t_{2}}{a {t_{1}}^{2} - a {t_{1}}^{2}} = 1$
$t_{1} + t_{2} = 2$ ..... $(3)$
$x = 2 a + a ({t_{1}}^{2} + t_{2}^{2} + t_{1} t_{2})$
(From $e q^{n}$ $(1)$ & $(2)$ )
$y = - a t_{1} t_{2} (t_{1} + t_{2})$

$t_{1} t_{2} = - \frac{y}{2 a}$

$x = 2 a + a ((t_{1} + t_{2})^{2} - t_{1} t_{2})$

$x = 2 a + a (4 + \frac{y}{2 a})$

$2 x - y - 12 a = 0$

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