Question

The eccentricity of the conic $\frac{{\left(x+2\right)}^2}{7}+{\left(y-1\right)}^2=1$ is

• A. $\sqrt{\frac{7}{8}}$
• B. $\sqrt{\frac{6}{17}}$
• C. $\sqrt{\frac{6}{7}}$
• D. $\sqrt{\frac{6}{11}}$

Answer 1

Answer: (C)

Given, $\frac{{\left(x+2\right)}^2}{7}+{\left(y-1\right)}^2=1$
$\Rightarrow \frac{{\left(x+2\right)}^2}{{\left(\sqrt{7}\right)}^2}+\frac{{\left(y-1\right)}^2}{{\left(1\right)}^2}=1$
It is an equation of ellipse
$\therefore a=\sqrt{7},b=1$
Eccentricity $=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{6}{7}}$