Question

The eccentricity of the ellipse $x^2+4y^2+2x+16y+13=0$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{2}$

$(x^2+2x)+4(y^2+4y)+13=0$

$\Rightarrow (x^2+2x+1)+4(y^2+4y+4)=4$

$\Rightarrow (x+1{)}^2+4(y+2{)}^2=4$

$\Rightarrow \frac{(x+2{)}^2}{4}+\frac{(y+2{)}^2}{1}=1$

Hence eccentricity $=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$