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If the normal to the parabola y^2=4a x a...

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.`

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A normal at point `t_(1)` cuts the parabola again at `t_(2)`. Then,
`t_(2)=-t_(1)-(2)/(t_(1))`
`or" "t_(1)^(2)+t_(1)t_(2)+2=0`
Since `t_(1)` is real, discriminant is greater than 0. Therefore,
`t_(2)^(2)-8ge0`
`or" "t_(2)^(2)ge8`
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