Which of the following lines, is a normal to the parabola $y^{2} = 16 x$ ?

y = x - 11 cos θ - 3 cos 3 θ

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y = x - 11 cos θ - cos 3 θ

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y = (x - 11) cos θ + cos 3 θ

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y = (x - 11) cos θ - cos 3 θ

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Solution

The correct option is B $y = (x - 11) cos θ - cos 3 θ$
$\begin{matrix} A c c o d i n g t o t h e q u e s t i o n, E q u o f p a r a b o l a i s . . . . . . . . y = m x - a m^{3} - 2 a m (w h e r e m = s l o p e, c o n d i t i o n o f n o r m a l i s, c = - 2 a m - a m^{3}) g i v e n, y = (x - 11) cos θ - cos 3 θ = x cos θ - 11 cos θ - cos 3 θ (w h e r e m = cos θ) c = - 11 cos θ - cos 3 θ = - 11 cos θ - (4 c o s^{3} θ - 3 cos θ) = - 8 cos θ - 4 {cos}^{3} θ i s t h e r e q u i r e d o p t i o n S o, t h e c o r r e c t o p t i o n i s D . \end{matrix}$

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