The normal at the point\[\left( {bt_1^2,2b{t_1}} \right)\] on a parabola meets the parabola again in the point \[\left( {bt_2^2,2b{t_2}} \right)\], then :
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Official solution
Equation of the normal to a parabola y2=4bx at point
\[\left( {bt_1^2,2b{t_1}} \right)\] is y=−t1x+2bt1+bt13
As given, it also passes through \[\left( {bt_2^2,2b{t_2}} \right)\] then
2bt2=t1bt22+2bt1+bt13
\[2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)\]
\[ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)\]
\[ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)\]
\[ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}\]
\[ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}\]