question_answer

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

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\[=\frac{1}{2}\left| \begin{matrix}    0 & 0 & 1  \\    a & -\frac{a}{2} & 1  \\    a & \frac{a}{2} & 1  \\ \end{matrix} \right|\]                 \[=\frac{1}{2}\left[ a\times \frac{a}{2}+a\times \frac{a}{2} \right]\]                 \[=\frac{{{a}^{2}}}{2}sq\,\,unit\]