Question

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

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

Answer 1

Answer: (B)

$x^2+2y^2-2x+3y+2=0$
$=(x-1{)}^2-1+2\left(y^2+\frac{3}{2}y+\frac{9}{16}-\frac{9}{16}\right)+2=0$
$=(x-1{)}^2+2{\left(y+\frac{3}{4}\right)}^2-\frac{9}{8}+1=0$
$=(x-1{)}^2+2{\left(y+\frac{3}{4}\right)}^2=\frac{1}{8}$
$=\frac{(x-1{)}^2}{\frac{1}{8}}+\frac{{\left(y+\frac{3}{4}\right)}^2}{\frac{1}{16}}=1$
Ellipse : $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{\frac{1}{16}}{\frac{1}{8}}}$
$e=\sqrt{1-\frac{8}{16}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$