The normal at the point (bt_1^2,2bt_1) on a parabola meets the parabola again at the point…

Question

The normal at the point $(bt_1^2,2bt_1)$ on a parabola meets the parabola again at the point $(bt_2^2,2bt_2)$. Then:

Accepted Answer

The normal to $y^2=4bx$ at $(bt_1^2,2bt_1)$ is $y=-t_1x+2bt_1+bt_1^3$. Since it also passes through $(bt_2^2,2bt_2)$:
$$2bt_2=-t_1(bt_2^2)+2bt_1+bt_1^3\ \Rightarrow\ 2(t_2-t_1)=-t_1(t_2^2-t_1^2)=-t_1(t_2-t_1)(t_2+t_1).$$
Dividing by $(t_2-t_1)\ne0$: $2=-t_1(t_2+t_1)\Rightarrow t_2=-t_1-\dfrac{2}{t_1}.$

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