Question

Find the lengths of the normals drawn from the point on the axis
of the parabola y = 8ax whose distance from the focus is 8a.

Answer 1

For y^2 =8ax

Focus is (2a,0)

Point on axis whose distance from focus is 8a is P (10a,0)

Equation of normal is

y=mx−2(2a)−(2a)m ^3

i.e y = mx-4a -2am^3

as it passes through P

so 0 = m(10a)-4a-2am^3

on solving we got , m = 0, ±√3

For m = 0,equation of normal is y =0

So the length of normal is 10a

and for m = √3

equation of normal is y = √3 x-4a√3 -2a(√3)^3

i.e y = √3 x - 10a√3

at y = 0

x = 10a

i.e N (10a, 0)..................(1)

Feet of normal is ( 2a(√3)^2, -4a√3)

i.e P (6a,-4a√3)..............(2)

Length of Normal PN = 8a from (1) & (2)

for m = -√3 the normal is symmetric so its length will also be 8a

So the lengths of normal are 10a, 8a, 8a.