Question

The area of the triangle formed by the lines ${\mathrm{x}}^2-{4\mathrm{y}}^2=0$ and $\mathrm{x}=\mathrm{a}$, is

• A. ${2\mathrm{a}}^2$
• B. $\frac{a^2}{2}$
• C. $\frac{\sqrt{3}{a}^2}{2}$
• D. $\frac{2a^2}{\sqrt{3}}$

Answer 1

The correct option is B

$\frac{a^2}{2}$

Explanation for the correct option:

Find the area of the triangle:

Given,

${\mathrm{x}}^2-4{\mathrm{y}}^2=0...\left(1\right)$

$\mathrm{x}=\mathrm{a}$

Equating equation (1),

$$\begin{gathered} {\mathrm{x}}^2=4{\mathrm{y}}^2 \\ \mathrm{x}=\pm 2\mathrm{y} \end{gathered}$$

Hence,

$\mathrm{x}=2\mathrm{y},\mathrm{x}=-2\mathrm{y}$

If $\mathrm{x}=\mathrm{a}$, then $\mathrm{a}=2\mathrm{y},\Rightarrow \mathrm{y}=\frac{\mathrm{a}}{2}$

If $\mathrm{x}=-\mathrm{a}$, then $-\mathrm{a}=2\mathrm{y},\Rightarrow \mathrm{y}=\frac{-\mathrm{a}}{2}$

If $\mathrm{x}=0$, then $0=2\mathrm{y},\Rightarrow \mathrm{y}=0$

Therefore the vertices of the triangle are,

$\left(0,0\right),\left(\mathrm{a},\frac{\mathrm{a}}{2}\right),\left(\mathrm{a},\frac{-\mathrm{a}}{2}\right)$

Applying area of triangle

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\times \left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\times \left[0\left(\frac{a}{2}-\left(-\frac{a}{2}\right)\right)+a\left(\left(-\frac{a}{2}\right)-0\right)+a\left(0-\frac{a}{2}\right)\right] \\ =\frac{1}{2}\times \left[0-\frac{a^2}{2}-\frac{a^2}{2}\right] \\ =\frac{1}{2}\times \left(-\frac{2a^2}{2}\right) \\ =-\frac{a^2}{2} \end{gathered}$$

Area of the triangle is $\frac{a^2}{2}$ . Since the area is always positive.

Hence, the correct answer is option (B).