Question

An equilateral triangle is inscribed in the parabola $y^2=4ax$, whose vertex is at the vertex of the parabola. The length of its side is

• A. $2a\sqrt{3}$
• B. $4a\sqrt{3}$
• C. $6a\sqrt{3}$
• D. $8a\sqrt{3}$

Answer 1

Answer: (D)

$In\Delta OCA$, $\tan 30^{\circ }=\frac{AC}{OC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{2at}{a{t}^2}$



$\Rightarrow t=2\sqrt{3}$

Again in $\Delta OCA$,
$OA=\sqrt{{\left(OC\right)}^2+{\left(CA\right)}^2}$

$=\sqrt{{\left(a{t}^2\right)}^2+{\left(2at\right)}^2}$

$=\sqrt{{\left\{a\cdot {\left(2\sqrt{3}\right)}^2\right\}}^2+{\left\{2a\cdot 2\sqrt{3}\right\}}^2}$

$=\sqrt{144a^2+48a^2}$

$=\sqrt{192a^2}$

$\Rightarrow OA=8\sqrt{3}a$