The feet of the normals to $y^{2} = 4 a x$ from the point $(6 a, 0)$ are

(0, 0)

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(4 a, 4 a)

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(4 a, - 4 a)

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(0, 0), (4 a, 4 a), (4 a, - 4 a)

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Solution

The correct option is C $(0, 0), (4 a, 4 a), (4 a, - 4 a)$
$\begin{matrix} y^{2} = 4 a x N : y = - t x + 2 a t + a t^{3} c o m e s f r o m (6 a, 0) \Rightarrow 0 = - t .6 a + 2 a t + a t^{3} \Rightarrow a t^{3} - 4 a t = 0 \Rightarrow a t (t^{2} - 4) = 0 \Rightarrow t = 0 a n d t^{2} - 4 = 0 t = \pm 2 a t t_{1} = 0 A (0, 0) A t t_{2} = 2 B (4 a, 4 a) a n d A t t_{3} = - 2 C (4 a, - 4 a) H e n c e, T h e o r d i n a t e s o f f o o t o f n o r m a l = 0, 4 a, - 4 a . \end{matrix}$
Hence, the option $D$ is the correct answer.

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