Question

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

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

Answer 1

The correct option is A $8\sqrt{3}a$

Parabola: $y^2=4ax......\left(i\right)$

let point $O$ parabola be $B$

Such that, $AB=BC=CA$

And as $ABC$ is equilateral triangle, $=>\angle BAC={60}^{\circ }$

$=>\angle BAS={30}^{\circ }$

line with slope $\frac{1}{\sqrt{3}}$ and passing through $(0,0)$ is

$y=\frac{1}{\sqrt{3}}x...............\left(ii\right)$

Point of intersection of $\left(i\right)$ and $\left(ii\right)=>\frac{1}{3}{x}^2=4ax$

$=>x=0,12a$ and $y=0,4\sqrt{3}a$

So, $B:(12a,4\sqrt{3}a)$ and $C:(12a,-4\sqrt{3}a)$

Length of side of triangle $=>AB=\sqrt{(12a-0{)}^2+{(4\sqrt{3}a-0)}^2}$

$=a\sqrt{144+48}$

$=a\sqrt{192}$

$=8\sqrt{3}a$.