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, $\left(x_{1}, y_{1}\right) \equiv\left(t_{1}^{2}, 2 t_{1}\right) \&\left(x_{2}, y_{2}\right) \equiv\left(t_{2}^{2}, 2 t_{2}\right)$
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, $\left(t_{1}-t_{2}\right)^{2}=4$
$\Rightarrow \left|t_{1}-t_{2}\right|=2$
$\Rightarrow \left|y_{1}-y_{2}\right|=4$