The normal drawn at a point $(2,-4)$ on the parabola $y^2=8 x$ cuts again the same parabola at $(\alpha,…
The normal drawn at a point $(2,-4)$ on the parabola $y^2=8 x$ cuts again the same parabola at $(\alpha, \beta)$ then $\alpha+\beta=$
$8$
$16$
$24$
$30$
Solution
Given the parabola, $y^2=8 x \Rightarrow a=2$
Now, given a point on $y^2=8 x$ is $(2,-4)$
$\Rightarrow(2,-4)=\left(a t^2, 2 a t\right) \Rightarrow(2,-4)=\left(2 t^2, 4 t\right) \Rightarrow t=-1$
Since, normal drawn at $(2,-4)$ again meet at $(\alpha, \beta)$
So, $(\alpha, \beta)=\left(2 t_2^2, 2 a t_2\right)$
where $t_2=-t-\frac{2}{t}=+1+2=3$
$\Rightarrow(\alpha, \beta)=\left(2 \times 3^2, 2 \times 2 \times 3\right)=(18,12)$
So, $\alpha+\beta=30$.