Given parabola $y^{2} = 4 a x$ , find the equation of Normal which will interest the normal at (8, 8) and the parabola at the same point.

2x + y = 24

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2x - y = 24

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x + y = 24

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2x + y + 24 = 0

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Solution

The correct option is A
2x + y = 24

It is useful to remember the fact that If normals at $t_{1} & t_{2}$ meet on the parabola then

$t_{1} t_{2} = 2$ .

Here $t_{1} = 2, a n d a = 2 & t_{1} t_{2} = 2$

$∴ t_{2} = 1$

We also know equation of normal for $y^{2} = 4 a x a t (a t^{2}, 2 a t)$ is $y = - t x + a t^{3} + 2 a t$ .

⇒ y = - 2x + 16 + 8

⇒ 2x + y = 24. is equation of the Normal.

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Find the equation of normal to the parabola $y^{2} = 8 x$ at (8, 8) using parametric form.

Given parabola $y^{2} = 4 a x$ , find the equation of Normal which will interest the normal at (8, 8) and the parabola at the same point.

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