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)

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