Q. Locus of mid point of any focal chord of $y^{2} = 4ax$ is
A) $y^{2} = a(x - 2a)$
B) $y^{2} = 2a(x - 2a)$
C) $y^{2} = 2a(x - a)$
D) None of these
Explanation
(c)
Let the mid point be $P(h,k).$ Equation of this chord is
$T = S_{1}\ ie,\ \ ky - 2a(x + h) = k^{2} - 4ah$
It must passes through $(a,\ 0)$
$( - 2a)(a + h) = k^{2} - 4ah$
Hence, the locus is $y^{2} = 2ax - 2a^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.