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Parametric Equation of Normal

Let P a t 2...

Question

Let $P (a t^{2}, 2 a t), Q (a r^{2}, 2 a r)$ be three points on a parabola $y^{2} = 4 a x$ . If $P Q$ is the focal chord and $P K, Q R$ are parallel where the co-ordinates of $K$ is $(2 a, 0)$ , then the value of $r$ is

A

\frac{t}{1 - t^{2}}

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B

\frac{1 - t^{2}}{t}

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C

\frac{t^{2} + 1}{t}

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D

\frac{t^{2} - 1}{t}

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Solution

The correct option is D $\frac{t^{2} - 1}{t}$
$m_{P K} = m_{Q R}$
$\frac{2 a t - 0}{a t^{2} - 2 a} = \frac{2 a t - 2 a r}{a {(t^{'})}^{2} - a r^{2}}$
$- t^{'} - t r^{2} = - t - r t^{2} - 2 t^{'} + 2 r, t t^{'} = - 1$
$t^{'} - t r^{2} = - t + 2 r - r t^{2}$
$- t r^{2} + r (t^{2} - 2) + t^{'} + t = 0$
$λ = \frac{(2 - t^{2}) \pm \sqrt{{(t^{2} - 2)}^{2} + 4 (- 1 + t^{2})}}{- 2 t} = \frac{2 - t^{2} \pm t^{2}}{- 2 t}$
$r = - \frac{1}{t}$
It is not possible as the $R$ and $Q$ will be one same
or $r = \frac{t^{2} - 1}{t}$

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