- #1

- 374

- 7

## Homework Statement

If the normal to the parabola y^2 = 4ax at the point (at^2 , 2at) cuts the parabola again at (aT^2, 2aT), then minimum value of T^2 is

ans: 8

I got the answer but I don't know why it should be the answer?

## Homework Equations

Equation of normal to the parabola in parametric form can be written as

y-2at = -t(x-at^2)

## The Attempt at a Solution

So, in the above equation, I substituted (aT^2, 2aT) as this point will lie on the normal.

2aT - 2at = -t(aT^2 - at^2)

On simplifying,

t^2 + tT + 2 = 0

From here on, I don't quite understand why what I did works.

Minimum value of this quadratic will be at (-b/2a, f(-b/2a) ).

i.e., (-T/2, f(-T/2) )

f(-T/2)= (- T^2 / 4 ) + 2

I equate ( -T^2 / 4 ) + 2 to 0.

I get T^2 = 8.

What did I just do?