The equation of the normal to the parabola, $x^{2} = 8 y$ at $x = 4$ is:

x + 2 y = 0

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x + y = 2

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x - 2 y = 0

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x + y = 6

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Solution

The correct option is B $x + y = 6$
Given equation of parabola-
$x^{2} = 8 y$

$∴ a = 2$

As we know that, parametric equation of normal to the parabola $x^{2} = 4 a y$ at t is-

$x + t y = 2 a t + a t^{3}$

$∴$ Equation of normal to the given parabola,

$x + t y = 4 t + 2 t^{3}$

Point of noramality - $(2 a t, t^{2})$

As we have to find normal at $x = 4$

$∴ 2 a t = 4$

$\Rightarrow 2 \times 2 \times t = 4$

$\Rightarrow t = 1$

$∴$ Equation of normal-

$x + y = 4 + 2$

$\Rightarrow x + y = 6$

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Similar questions

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

Equation of the common tangent to the parabola $y^{2} = 24 x$ and the circle $x^{2} + y^{2} = 18$ is

x^{2} + y^{2} - 2 x + 4 y = 0

y^{2} = 4 x

x y = 2

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