Question

The eccentricity of the conic represented by the equation $x^2+2y^2-2x+3y+2=0$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

$x^2+2y^2-2x+3y+2=0$
$\Rightarrow (x-1{)}^2+2(y+\frac{3}{4}{)}^2=\frac{1}{8}$
$\Rightarrow \frac{(x-1{)}^2}{1/8}+\frac{(y+3/4{)}^2}{1/16}=1$
It is an ellipse with $a^2=1/8,b^2=1/16$ .Hence its eccentricity
$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{8}{16}}=\frac{1}{\sqrt{2}}$