Question

An equilateral triangle is increased in parabola y2 = 8x whose one vertex coincides with vertex of parabola. Find the area of triangle.

Answer 1

The parabola $y^2=8x$ is a rightward parabola and its vertex is origin O(0, 0).

Let OAB be the triangle inscribed in the parabola.

Assume the edge AB is parallel to the directrix, so that the vertices of the triangle are O(0, 0), A(x, y) and B(x, -y).

Length of AB $=2y.$

The axis of the parabola bisects AB and also $\angle AOB$ since the triangle is equilateral. Let the point of bisection, i.e., midpoint of AB, be C.

Since triangle ABC is equilateral, $\angle AOB=60^o.$

Since the axis bisects this angle,

$\longrightarrow\angle AOC=\angle BOC=30^o$

Here triangle AOC is a right triangle, right angled at A, so we have,

$\longrightarrow \tan30^o=\dfrac{AC}{OC}$

$\longrightarrow\dfrac{1}{\sqrt3}=\dfrac{y}{x}$

$\longrightarrow x=y\sqrt3$

From equation of parabola,

$\longrightarrow y^2=8y\sqrt3$

Since $y$ is non - zero,

$\longrightarrow y=8\sqrt3$

And so,

$\longrightarrow x=(8\sqrt3)\sqrt3$

$\longrightarrow x=24$

Hence area of triangle OAB is,

$\longrightarrow\dfrac{1}{2}\times OC\times AB=\dfrac{1}{2}\,x(2y)$

$\longrightarrow\dfrac{1}{2}\times OC\times AB=xy$

$\longrightarrow\underline{\underline{\dfrac{1}{2}\times OC\times AB=192\sqrt3}}$