A) 0
B) 1/2
C) \[1/\sqrt{2}\]
D) \[\sqrt{2}\]
Correct Answer: C
Solution :
Equation \[{{x}^{2}}+2{{y}^{2}}-2x+3y+2=0\] can be written as \[\frac{{{(x-1)}^{2}}}{2}+{{\left( y+\frac{3}{4} \right)}^{2}}=\frac{1}{16}\]Þ\[\frac{{{(x-1)}^{2}}}{(1/8)}+\frac{{{\left( y+\frac{3}{4} \right)}^{2}}}{(1/16)}=1\], which is an ellipse with \[{{a}^{2}}=\frac{1}{8}\] and \[{{b}^{2}}=\frac{1}{16}\] \[\therefore \frac{1}{16}=\frac{1}{8}(1-{{e}^{2}})\] Þ\[{{e}^{2}}=1-\frac{1}{2}\]Þ\[e=\frac{1}{\sqrt{2}}\].
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