Question

The normal to the parabola $y^2=8x$ at the point $P(2,4)$ meets it again at the point $Q(l,m)$. If the normal to the parabola at $Q$ meets it again at $R(\alpha ,\beta )$, then the value of $9\alpha +6\beta +\frac{l}{9}+\frac{m}{6}$ is equal to

Answer 1

Given parabola is $y^2=8x(a=2)$
$\Rightarrow$ Any point on the parabola will be in the form of $(2t^2,4t)$
Let $P(2,4)=(2t_1^2,4t_1)\Rightarrow t_1=1$

Let normal at $P$ meets the parabola at $Q(2t_2^2,4t_2).$
$\therefore t_2=-t_1-\frac{2}{t_1}=-1-2=-3$
$\Rightarrow Q(2t_2^2,4t_2)=Q(18,-12)\equiv Q(l,m)$

Let normal at $Q$ meets the parabola at $R(2t_3^2,4t_3)$.
$\therefore t_3=-t_2-\frac{2}{t_2}=3+\frac{2}{3}=\frac{11}{3}$
$\Rightarrow R(2t_3^2,4t_3)=R(\frac{242}{9},\frac{44}{3})\equiv R(\alpha ,\beta )$

$\therefore 9\alpha +6\beta +\frac{l}{9}+\frac{m}{6}=242+88+2-2=330$