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Q.
The eccentricity of the curve $2 x ^{2}+ y ^{2}-8 x -2 y +1=0$ is:
Conic Sections
Solution:
The given curve is: $2 x^{2}-8 x+y^{2}-2 y+1=0$
$\Rightarrow 2\left(x^{2}-4 x+4-4\right)+\left(y^{2}-2 y+1\right)=0$
$\Rightarrow 2(x-2)^{2}-8+(y-1)^{2}=0$
$\Rightarrow 2(x-2)^{2}+(y-1)^{2}=8 $
$\Rightarrow \frac{(x-2)^{2}}{4}+\frac{(y-1)^{2}}{8}=1$
This is equation of ellipse with centre $(2,1)$
$\Rightarrow a ^{2}=4, b ^{2}=8$
Eccentricity $e =\sqrt{\frac{ b ^{2}- a ^{2}}{ b ^{2}}}=\sqrt{\frac{8-4}{8}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$