Question

Find the length of the side of an equilateral triangle inscribed in the parabola, $y^2=4x$ so that one of its angular point is at the vertex.

• A. $16\sqrt{3}$
• B. $4\sqrt{3}$
• C. $8\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is C $8\sqrt{3}$

Let $△APQ$ be the equilateral triangle inscribed symmetrically w.r.t to the axis of the parabola $y^2=4x$
Since the triangle is placed symmetrically inside the parabola
$\angle PAM=\angle QAM={30}^0$
Let the length of the side AP of the triangle APQ be r and coordinate of the point p be $(h,k)$
$\therefore h=AM=r\cos {30}^0=\frac{\sqrt{3}}{2}r$
$k=MP=r\sin {30}^0=\frac{1}{2}$
Hence the coordinate of the point P are $(\frac{\sqrt{3}}{2}r,\frac{1}{2}r)$
Since $P(\frac{\sqrt{3}}{2}r,\frac{r}{2})$ lies on the parabola $y^2=4ax$ we get
${\left(\frac{r}{2}\right)}^2=4\left(\frac{\sqrt{3}}{2}r\right)\Rightarrow r=8\sqrt{3}$