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Equation of Normal at Given Point

Normal equati...

Question

Normal equations to parabola $y^{2} = 4 a x$ passing through point $(5 a, 2 a)$ are

y = - 3 x + 17 a

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x = - 3 y + 11 a

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y = x - 3 a

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y = - 2 x + 12 a

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Solution

The correct option is D $y = - 2 x + 12 a$
Normal to $y^{2} = 4 a x$ is
$y = - t x + a t^{3} + 2 a t$
Since it passes through $(5 a, 2 a)$
$∴ 2 a = - 5 a t + a t^{3} + 2 a t \Rightarrow 2 a = a t^{3} - 3 a t \Rightarrow t^{3} - 3 t - 2 = 0 (∵ a \neq 0)$
By observation $t = - 1$ is one root,
$\Rightarrow (t + 1) (t^{2} - t - 2) = 0 \Rightarrow (t + 1)^{2} (t - 2) = 0 ∴ t = - 1 or t = 2$
So, the equation of normal are
$t = - 1 \Rightarrow y = x - a - 2 a \Rightarrow y = x - 3 a t = 2 \Rightarrow y = - 2 x + 8 a + 4 a \Rightarrow y = - 2 x + 12 a$

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