Q. Eccentricity of the ellipse $x^{2} + 2y^{2} - 2x + 3y + 2 = 0$ is
A) $\frac{1}{\sqrt{2}}$
B) $\frac{1}{2}$
C) $\frac{1}{2\sqrt{2}}$
D) $\frac{1}{\sqrt{3}}$
Explanation
(a)
Given equation of ellipse can be rewritten as
$\frac{(x - {1)}^{2}}{1/8} + \frac{\left( y + \frac{3}{4} \right)^{2}}{1/16} = 1$
$\therefore Eccentricity = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{8}{16}} = \frac{1}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.