If two normals to a parabola $y^{2} = 4 a x$ intersect at right angles, then the chord joining their feet passes through a fixed point whose co-ordinates are

(- 2 a, 0)

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(a, 0)

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(2 a, 0)

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none of these

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Solution

The correct option is C $(a, 0)$
Given the equation of parabola is
$y^{2} = 4 a x$

Let $P (a {t_{1}}^{2}, 2 a t_{1})$ and $Q (a {t_{1}}^{2}, 2 a t_{1})$ be two points on the parabola.

Equation of normal at $P$ is given as
$y = - t_{1} x + 2 a t_{1} + a {t_{1}}^{3}$

Slope of normal at point $P$ is $- t_{1}$

Equation of normal at $Q$ is given as
$y = - t_{2} x + 2 a t_{2} + a {t_{2}}^{3}$

Slope of normal at point $Q$ is $- t_{2}$ .

Equation of chord joining $P (a {t_{1}}^{2}, 2 a t_{1})$ and $Q (a {t_{1}}^{2}, 2 a t_{1})$ is

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

or $2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$

But $t_{1} t_{2} = - 1$

Chord $P Q$ is $2 x - (t_{1} + t_{2}) y - 2 a = 0$

or $(2 x - 2 a) - (t_{1} + t_{2}) y = 0$

or $y = \frac{2 (x - a)}{(t_{1} + t_{2})}$

$\Rightarrow$ Chord $P Q$ passes through the fixed point $(a, 0)$ .

Suggest Corrections

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