Question

The eccentricity of the curve $2x^2+y^2-8x-2y+1=0$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

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