Question
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-
A.
$${t_2} = {t_1} + \frac{2}{{{t_1}}}$$
B.
$${t_2} = - {t_1} - \frac{2}{{{t_1}}}$$
C.
$${t_2} = - {t_1} + \frac{2}{{{t_1}}}$$
D.
$${t_2} = {t_1} - \frac{2}{{{t_1}}}$$
Answer :
$${t_2} = - {t_1} - \frac{2}{{{t_1}}}$$
Solution :
Equation of the normal to a parabola $${y^2} = 4bx$$ at point $$\left( {bt_1^2,\,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$
As given, it also passes through $$\left( {bt_2^2,\,2b{t_2}} \right)$$ then
$$\eqalign{
& 2b{t_2} = - {t_1}bt_2^2 + 2b{t_1} + bt_1^3 \cr
& 2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right) \cr
& \Rightarrow 2{t_2} - 2{t_1} = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right) \cr
& \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right) \cr
& \Rightarrow {t_2} + {t_1} = - \frac{2}{{{t_1}}} \cr
& \Rightarrow {t_2} = - {t_1} - \frac{2}{{{t_1}}} \cr} $$