An equilateral triangle is inscribed in the parabola $y^2=8 x$, with one of its vertices is the vertex of…

An equilateral triangle is inscribed in the parabola $y^2=8 x$, with one of its vertices is the vertex of the parabola. Then, length of the side of that triangle is
  1. $24 \sqrt{3}$ units
  2. $16 \sqrt{3}$ units
  3. $8 \sqrt{3}$ units
  4. $4 \sqrt{3}$ units

Solution

Let $a$ be the length of the side of an equilateral triangle.
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

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