Find the equation of normal to the parabola
$y^{2} = 16 x$ at point (4,8)

x + y + 4 = 0

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x + y - 2 = 0

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x + y - 3 = 0

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x + y - 4 = 0

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Solution

The correct option is D
x + y - 4 = 0

For a given point (h,k) on the parabola $y^{2} = 4 a x$ the equation of normal at that point is

$y - k = \frac{- k}{2 a (x - h)}$

Here (h,k) = (4,8) and a = 4.

Therefore equation of normal is
$y - 8 = \frac{- 8}{2} (4) (x - 4)$

$x + y - 4 = 0$

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