If an equilateral
is inscribed in a parabola ${\mathrm{y}}^2=12\mathrm{x}$ with one of the vertex is at the vertex of the parabola then its height is

Given parabola

$$\begin{gathered} {\mathrm{y}}^2=12\mathrm{x} \\ \mathrm{a}=3, \end{gathered}$$ 

$$\begin{gathered} Here AOB is an equilateral triangle inscribed in parabola \\ then A=(a\cos {30}^0,a\sin {30}^0) \\ =(\frac{\sqrt{3}}{2}a,\frac{a}{2}) \\ Since A lies on parabola then {\left(\frac{a}{2}\right)}^2=12\left(\frac{\sqrt{3}}{2}a\right) \\ \Rightarrow \frac{a^2}{4}=6\sqrt{3}a \\ \Rightarrow a=24\sqrt{3} \\ length of side of equilateral triangle=24\sqrt{3} \end{gathered}$$

  Now height of an equilateral triangle    $=\frac{\sqrt{3}}{2}\times \mathrm{side}$

                $$\begin{gathered} =\frac{\sqrt{3}}{2}\times 24\sqrt{3} \\ =36 \end{gathered}$$