Question

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

Answer 1

Draw diagram and assume points

Given: $y^2=4ax$ $\dots \dots \dots \left(1\right)$

Let, point $A\text{and}B$ are $(a{t}^2,2at)\text{and}(a{t}^2,-2at)$ respectively.

$AB=CA+CB=2at+2at=4at$

Using concept of equilateral triangle

$\because △OAB$ is an equilateral triangle, $OA=AB$,
Using distance formula,

$\sqrt{(a{t}^2-0{)}^2+(2at-0{)}^2}=4at$

Squaring on both sides

$\Rightarrow (a{t}^2{)}^2+(2at{)}^2=(4at{)}^2$

$\Rightarrow a^2{t}^4+4a^2{t}^2=16a^2{t}^2$

Divide by $a^2$ on both sides

$\Rightarrow t^4+4t^2=16t^2$

$\Rightarrow t^4-12t^2=0$

$\Rightarrow t^2(t^2-12)=0$

$\Rightarrow t^2=0,12$

$\because t$ cannot be $0,t^2=12$

$\because t=\pm 2\sqrt{3}$

Finding length of side of triangle.

Thus, $AB=4at$

$=4\times a\times 2\sqrt{3}$
$=8\sqrt{3}a$

Hence, the side of the equilateral triangle inscribed in parabola $y^2=4ax$ is $=8\sqrt{3}a$