Question Id : 17914 |
Context : JEE Main 2026 (21 January Evening Shift)
Let $y^2 = 12x$ be the parabola with its vertex at $O$. Let $P$ be a point on the parabola and $A$ be a point on the y-axis such that $OPA = 90^\circ$. Then the locus of the centroid of triangle $OPA$ is:
🎥 Video solution / Text Solution of this question is given below:
Let $P(3t^2, 6t)$
$m_{AP} = \frac{t}{2}$
Equation of $AP$ is
$y - 6t = \frac{t}{2}(x - 3t^2)$
Put $y = 0 \Rightarrow x = 12 + 3t^2$
$\Rightarrow A(12 + 3t^2, 0)$
Let centroid of $\triangle OPA$ be $(h,k)$
$\Rightarrow 3h = 0 + 3t^2 + 12 + 3t^2$
$\Rightarrow 3k = 0 + 6t + 0$
$\Rightarrow t = \frac{k}{2},; h = 2t^2 + 4$
$\Rightarrow h = \frac{k^2}{2} + 4$
$\Rightarrow$ locus of $(h,k)$ is
$y^2 = 2x - 8$