Question

An equilateral triangle is inscribed in the parabola $y^2=16ax$ with one of its vertices at the origin. Then, the centroid of that triangle is

• A. (8 a, 0)
• B. (16 a, 0)
• C. (32 a, 0)
• D. (48 a, 0)

Answer 1

Answer: (C)

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 $y^2=16ax$
$\frac{l^2}{4}=16a\frac{\sqrt{3}}{2}l$
$\Rightarrow l=\frac{16\sqrt{3}}{2}\cdot 4a$
$\Rightarrow 32\sqrt{3}a$
$\Rightarrow A=(48a,16\sqrt{3}a)O(0,0)B(48a,-16\sqrt{3})$
Centroid $=\left(\frac{48a+48a+0}{3},\frac{16\sqrt{3a}-16\sqrt{3a}+0}{3}\right)$
$=(32a,0)$