A) \[\sqrt{\frac{2}{3}}\]
B) \[\sqrt{\frac{3}{2}}\]
C) \[{{e}^{1/2}}\]
D) None of these
Correct Answer: A
Solution :
We have,\[{{x}^{2}}-4x+4{{y}^{2}}=12\] \[\Rightarrow \]\[{{(x-2)}^{2}}+4{{(y-0)}^{2}}=8\] \[\Rightarrow \]\[\frac{{{(x-2)}^{2}}}{{{(2\sqrt{2})}^{2}}}+\frac{{{(y-0)}^{2}}}{(\sqrt{2})}=1\] which is an elhpse whose major and minor axes are \[2a=2\sqrt{2}\] and \[2b=\sqrt{2}\]repectively. \[\therefore \]Its eccentricity e is given by \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\] \[=\sqrt{1-\frac{2}{8}}=\frac{\sqrt{3}}{2}\]
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