Question

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

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

Answer 1

Answer: (B)

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

Given lines are $x^2-4y^2=0$
$\Rightarrow (x-2y)(x+2y)=0$
$\Rightarrow (x-2y)=0,(x+2y)=0$ and $x=a$
On drawing these lines, we get the following triangle.
Here, $A(0,0),B(a,-\frac{a}{2})$ and $C(a,\frac{a}{2})$ are the vertices of triangle.
Therefore, required area $=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ a& -\frac{a}{2}& 1\\ a& \frac{a}{2}& 1\end{array}\right|$
$=\frac{1}{2}[a\times \frac{a}{2}+a\times \frac{a}{2}]$
$=\frac{a^2}{2}$ sq unit