If the normal to the parabola `y^2=4a x` at point `t_1` cuts the parabola again at point `t_2` , then prove that `t2 2geq8.`

Explanation

To solve the problem, we need to prove that \( t_2^2 \geq 8 \) given that the normal to the parabola \( y^2 = 4ax \) at point \( t_1 \) intersects the parabola again at point \( t_2 \).

### Step-by-Step Solution:

1. **Identify the Point on the Parabola**:
The point on the parabola \( y^2 = 4ax \) corresponding to the parameter \( t_1 \) is given by:
\[
P(t_1) = (at_1^2, 2at_1)
...