A) \[2{{a}^{2}}\]
B) \[\frac{{{a}^{2}}}{2}\]
C) \[\frac{\sqrt{3}{{a}^{2}}}{2}\]
D) \[\frac{2{{a}^{2}}}{\sqrt{3}}\]
Correct Answer: B
Solution :
Given lines are \[{{x}^{2}}-4{{y}^{2}}=0\] \[\Rightarrow\] \[(x-2y)(x+2y)=0\] \[\Rightarrow \] \[(x-2y)=0,\,\,(x+2y)=0\]and\[x=a\] On drawing these lines, we get the following triangle Here, \[A(0,\,\,0),\,\,B\left( a,\,\,-\frac{a}{2} \right)\]and\[C\left( a,\,\,\frac{a}{2} \right)\]are the vertices of triangle. \[\therefore \]Required area\[=\frac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ a & -\frac{a}{2} & 1 \\ a & \frac{a}{2} & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}\left[ a\times \frac{a}{2}+a\times \frac{a}{2} \right]\] \[=\frac{{{a}^{2}}}{2}sq\,\,unit\]You need to login to perform this action.
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