Question

The area (in sq. units) of an equilateral triangle inscribed in the parabola $y^2=8x$, with one of its vertices on the vertex of this parabola,

• A. $128\sqrt{3}$
• B. $192\sqrt{3}$
• C. $64\sqrt{3}$
• D. $256\sqrt{3}$

Answer 1

The correct option is B

$192\sqrt{3}$

Explanation for Correct Answer:

Determining the area of the equilateral triangle.

The graph of the given problem is as shown:

Let $ABO$is an equilateral triangle with side equal to ‘$a$’.

Now, a component of $OA$along positive $x$-axis is $a\cos (30°)$

Also, a component of $OA$along positive $y$-axis is $a\sin (30°)$

Hence, the coordinates of a point $A$are,

$A(a\cos \left(30°\right),a\sin \left(30°\right))$ which lies on the parabola.

Now $y^2=8x$

Put $y=a\sin (30°)$ and $x=a\cos (30°)$ in a parabolic equation we get,

$$\begin{gathered} \Rightarrow {\left(a\sin (30°)\right)}^2=8\times \left(a\cos \left(30°\right)\right) \\ \Rightarrow a^2\times {\left(\frac{1}{2}\right)}^2=8\times a\times \left(\frac{\sqrt{3}}{2}\right)\left[\because \sin (30°)=\frac{1}{2},\cos \left(30°\right)=\frac{\sqrt{3}}{2}\right] \\ \Rightarrow \frac{a}{4}=4\times \sqrt{3} \\ \Rightarrow a=16\sqrt{3} \end{gathered}$$

Now area of an equilateral triangle is equal to

$$\begin{gathered} Ar(△)=\frac{\sqrt{3}}{4}{a}^2 \\ =\frac{\sqrt{3}}{}\times {\left(16\sqrt{3}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 768 \\ =192\sqrt{3} \end{gathered}$$

Therefore the area of an equilateral triangle is equal to $192\sqrt{3}$ square units.

Hence, option (B) is correct