Thank you for reporting, we will resolve it shortly
Q.
An equilateral triangle is inscribed in the parabola $y^{2}=16\, ax$ with one of its vertices at the origin. Then, the centroid of that triangle is
TS EAMCET 2018
Solution:
Let the length at one side is $l$, so coordinate of $A$ is
$\left(\frac{\sqrt{3} l}{2}, \frac{l}{2}\right)$ lies on parabola so in $y^{2}=16 ax$
$\frac{l^{2}}{4}=16 a \frac{\sqrt{3}}{2} l$
$\Rightarrow \, l=\frac{16 \sqrt{3}}{2} \cdot 4 a$
$ \Rightarrow \,32 \sqrt{3} a$
$\Rightarrow \,A=(48 a, 16 \sqrt{3} a) O(0,0) B(48 a,-16 \sqrt{3})$
Centroid $=\left(\frac{48 a+48 a+0}{3}, \frac{16 \sqrt{3 a}-16 \sqrt{3 a}+0}{3}\right) $
$=(32 \,a, 0)$
