Question

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. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

Step: 1 Simplify the given equation into a general equation of ellipse.

$\begin{array}{rcl}{x}^2+2y^2-2x+3y+2& =& 0\end{array}$

$\Rightarrow$$\begin{array}{rcl}& & (x^2-2\times 1\times x+1)+2[y^2+2\times y\times \frac{3}{4}+{\left(\frac{3}{4}\right)}^2]+1-\frac{9}{8}\end{array}=0\begin{array}{rcl}& & \end{array}$

$\Rightarrow$ $\begin{array}{rcl}{(x-1)}^2+2{(y+\frac{3}{4})}^2& =& \frac{1}{8}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{{(x-1)}^2}{{\displaystyle \frac{1}{8}}}+\frac{{(y+\frac{3}{4})}^2}{{\displaystyle \frac{1}{16}}}& =& 1\end{array}$

The general equation of ellipse is $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$

If we compare this equation, then we will get:

$a^2=\frac{1}{8},b^2=\frac{1}{16}$.

Step2. Finding eccentricity.

Therefore, eccentricity$\left(e\right)$

$\begin{array}{rcl}e& =& \sqrt{1-\frac{b^2}{a^2}}\\ & =& \sqrt{1-\frac{\left({\displaystyle \frac{1}{16}}\right)}{\left({\displaystyle \frac{1}{8}}\right)}}\\ & =& \sqrt{1-\frac{8}{16}}\\ & =& \sqrt{\frac{1}{2}}\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathit{e}& \mathbf{=}& \frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\end{array}$

Hence, Option (A) is the correct answer.