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 `(t_2)^2geq8.`

Explanation

To prove that \( (t_2)^2 \geq 8 \) for the normal to the parabola \( y^2 = 4ax \) at point \( t_1 \) that intersects the parabola again at point \( t_2 \), we will follow these steps:

### Step 1: Identify the coordinates of the point on the parabola
The point on the parabola corresponding to \( t_1 \) is given by:
\[ P(t_1) = (at_1^2, 2at_1) \]

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