Question

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

• A. x + y + 4 = 0
• B. x + y - 2 = 0
• C. x + y - 3 = 0
• D. x + y - 4 = 0

Answer 1

The correct option is D

x + y - 4 = 0

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

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

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

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

$x+y-4=0$