A) \[{{X}^{2}}-4{{Y}^{2}}={{a}^{2}}\]
B) \[{{X}^{2}}+4{{Y}^{2}}={{a}^{2}}\]
C) \[{{X}^{2}}+4{{Y}^{2}}=-{{a}^{2}}\]
D) None of these
Correct Answer: A
Solution :
Here,\[\theta =2\], So, \[\cos \theta =\frac{1}{\sqrt{5}},\,\,\sin \theta =\frac{2}{\sqrt{5}}\] For\[x\]and\[y\], we have \[x=X\cos \theta -Y\sin \theta =\frac{X-2Y}{\sqrt{5}}\] and \[y=X\sin \theta +Y\cos \theta =\frac{2X+Y}{\sqrt{5}}\] The equation\[4xy-3{{x}^{2}}={{a}^{2}}\]reduces to \[\frac{4(X-2Y)}{\sqrt{5}}\cdot \frac{2(X+Y)}{\sqrt{5}}-3{{\left( \frac{X-2Y}{\sqrt{5}} \right)}^{2}}={{a}^{2}}\] \[\Rightarrow \] \[4(2{{X}^{2}}-2{{Y}^{2}}-3XY)\] \[-3({{X}^{2}}-4XY+4{{Y}^{2}})=5{{a}^{2}}\] \[\Rightarrow \] \[5{{X}^{2}}-20{{Y}^{2}}=5{{a}^{2}}\] \[\Rightarrow \] \[{{X}^{2}}-4{{Y}^{2}}={{a}^{2}}\]
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