Q. Let $O$ be the origin and $A$ be a point on the curve $y^{2} = 4x.$ Then, the locus of the mid point of $OA,$ is
A) $x^{2} = 4y$
B) $x^{2} = 2y$
C) $x^{2} = 16y$
D) $y^{2} = 2x$
Explanation
(d)
Let coordinates of $O$ and $A$ (0, 0 ) and $\left( at^{2},\ 2at \right)$ respectively
$\therefore$ Coordinates of mid point of$\ OA\$ are
$\left( \frac{0 + at^{2}}{2},\frac{0 + 2at}{2} \right) = \left( \frac{at^{2}}{2},\ at \right)$
Since , $\left( at^{2} \right) = 4\ \left( \frac{at^{2}}{2} \right)$
Hence, that locus of required point is $y^{2} = 2x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.