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 $4 x^{2}+y^{2}-8 x+4 y-8=0$ 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 $(e)=\sqrt{1-\frac{a^{2}}{b^{2}}}$
$=\sqrt{1-\frac{4}{16}}=\sqrt{\frac{12}{16}}=\frac{\sqrt{3}}{2}$