Question

If the normals at two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ of the parabola $y^2=4x$ meets again on the parabola, where $x_1+x_2=8,$ then $\left|y_1-y_2\right|$ is equal to

• A. $\sqrt{2}$
• B. $3$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

Let, $\left(x_1,y_1\right)\equiv \left(t_1^2,2t_1\right)\&\left(x_2,y_2\right)\equiv \left(t_2^2,2t_2\right)$
Now, $x_1+x_2=8$ and both the normals again meets on the parabola,
$\therefore t_1^2+t_2^2=8\&t_1{t}_2=2$
On solving, we get, ${\left(t_1-t_2\right)}^2=4$
$\Rightarrow \left|t_1-t_2\right|=2$
$\Rightarrow \left|y_1-y_2\right|=4$