Find the ordinate of the feet of three normals to the parabola $y^{2} = 4 x$ from the point $(6 a, 0)$

0. - 3 a, 3 a

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0, - 2 a, 2 a

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0, - 4 a, 4 a

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0. - 5 a, 5 a

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Solution

The correct option is C $0, - 4 a, 4 a$
We get
$y^{2} = 4 a x$
$N : y = - t x + 2 a t + a t^{3}$
Comes from $(6 a, 0)$
$0 = - t : 6 a + 2 a t + a t^{3}$
$a t^{3} - 4 a t = 0$
$a t (t^{2} - 4) = 0$
$t = 0 t^{2} - 4 = 0$
$t = \pm 2$
$t_{1} = 0$ $t_{2} = 2$ $t_{3} = - 2$
$A (0, 0)$ $B (4 a, 4 a)$ $C (4 a, - 4 a)$
Ord. of total of normals $= 0, 4 a, - 4 a$
Hence,
Option $(C)$ is correct answer

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