Question

Eccentricity of the ellipse $4x^2+y^2-8x+4y-8=0$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{\sqrt{3}}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{8}$
• E. $\frac{\sqrt{3}}{16}$

Answer 1

Answer: (A)

Equation $4x^2+y^2-8x+4y-8=0$ is an ellipse.
$\Rightarrow 4(x-1{)}^2+(y+2{)}^2-8-8=0$
$=4(x-1{)}^2+(y+2{)}^2=16$
$=\frac{(x-1{)}^2}{4}+\frac{(y+2{)}^2}{16}=1$, where $b>a$
$\therefore$ Eccentricity $(e)=\sqrt{1-\frac{a^2}{b^2}}$
$=\sqrt{1-\frac{4}{16}}=\sqrt{\frac{12}{16}}=\frac{\sqrt{3}}{2}$