The normals to the parabola $y^{2} = 4 a x$ from the point $(5 a, 2 a)$ are

y = 3 x + 33 a

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x = 3 y + 3 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 options are
C $y = - 2 x + 12 a$
D $y = x - 3 a$
Equation of normal at point $t$ to the parabola $y^{2} = 4 a x$ is given by
$y + t x = 2 a t + a t^{3}$
Given it passes through $(5 a, 2 a)$
$\Rightarrow 2 a + 5 a t = 2 a t + a t^{3} \Rightarrow t^{3} - 3 t - 2 = 0 \Rightarrow (t + 1)^{2} (t - 2) = 0 \Rightarrow t = - 1, 2$
Hence required normals are
$y = x - 3 a, y = - 2 x + 12 a$

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