Question

An equilateral triangle is inscribed in the parabola $y^2=4ax$ such that one vertex of this triangle coincides with the vertex of the parabola. The length of side of this triangle is

• A. $2a\sqrt{3}$
• B. $4a\sqrt{3}$
• C. $6a\sqrt{3}$
• D. $8a\sqrt{3}$

Answer 1

The correct option is D $8a\sqrt{3}$

Here, parabola lies at origin.
Let $△OAB$ be an equilateral triangle.
$\Rightarrow \angle AOB=\frac{\pi }{3}$
$\therefore$ perpendicular distance of $O$ to $AB$
$=k\times \sin \frac{\pi }{3}=k\times \frac{\sqrt{3}}{2}$
$=x-$ coordinate of $A$

Similarly, $y-$coordinate of $A$
$=k\cos \frac{\pi }{3}=\frac{k}{2}$

Since, point $A$ lies on parabola $y^2=4ax$

so, ${\left(\frac{k}{2}\right)}^2=4a\frac{\sqrt{3}}{2}k$
$\Rightarrow k=8a\sqrt{3}$

Here, the length of the triangle is $2\times \frac{k}{2}=k$
$\therefore$ length of the triangle $8a\sqrt{3}$