An equilateral triangle is inscribed in the parabola $y^2=8 x$, with one of its vertices is the vertex of…
- $24 \sqrt{3}$ units
- $16 \sqrt{3}$ units
- $8 \sqrt{3}$ units
- $4 \sqrt{3}$ units
Solution

Then, from above figure, we can say that the point $\left(\frac{\sqrt{3}}{2} a, \frac{a}{2}\right)$ will lie on parabola $y^2=8 x$. So, $\left(\frac{a}{2}\right)^2=8\left(\frac{\sqrt{3}}{2} a\right) \Rightarrow a^2=16 \sqrt{3} a \Rightarrow a=16 \sqrt{3}$ units
Asked in: AP EAMCET 2015