Question

The point of intersection of normals to the parabola $y^2=4x$ at the points whose ordinates are $4$ and $6$ is

• A. $(30,-21)$
• B. $(21,-30)$
• C. $(17,-19)$
• D. $(19,-17)$

Answer 1

The correct option is B $(21,-30)$

Parabola is $y^2=4x$
let the points be $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$.
$a=1$ then points will be $P(t_1^2,2t_1)$ and $Q(t_2^2,2t_2)$
Normals are drawn at $P$ and $Q$
Given that ordinates of $P$ and $Q$ are $4$ and $6$ so,
$2t_1=4\Rightarrow t_1=2$ and $2t_2=6\Rightarrow t_2=3$
Now, point of intersection of normals at $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ is $R\equiv (2a+a(t_1^2+t_2^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2\left)\right)$
substitute $t_1=2$ and $t_2=3,a=1$
$R\equiv (21,-30)$