The feet of the normals to y2 = 4ax drawn from (6a,0) are | MATH - PARABOLA | SolveeIt | Solveeit

Today, August 19

Question

The feet of the normals to y² = 4ax drawn from (6a,0) are

A

(0,0) (4a, 4a) (4a,-4a)

B

(0,0) (a,2a) (a, -2a)

C

(0,0) (6a, 9a) (6a, -9a)

D

(0, 0) (a, a) (-a, a)

Answer

(0,0) (4a, 4a) (4a,-4a)

Explanation

Solution

The equation of Normal at ‘t’ is y+xt = 2at + at³, this passes through (6a,0).

⇒ 0 + 6at = 2at + at³

⇒ t = 0, 2, -2

∴ The corresponding points are (0, 0) (4a, 4a) (4a,-4a)