Question

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

• A. $24\sqrt{3}$
• B. $16\sqrt{3}$
• C. $36$
• D. $24$

Answer 1

The correct option is C $36$


coordinates of point $A\equiv (l\cos {30}^{\circ },l\sin {30}^{\circ })$
$A\equiv (\frac{l\sqrt{3}}{2},\frac{l}{2})$

Point lies on parabola so,
${\left(\frac{l}{2}\right)}^2=12\cdot \left(\frac{l\sqrt{3}}{2}\right)$
$\Rightarrow l=24\sqrt{3}$

Height of this triangle $OC=l\cos {30}^{\text{o}}=l\cdot \frac{\sqrt{3}}{2}=24\sqrt{3}\cdot \frac{\sqrt{3}}{2}=36$