Q. The eccentricity of the conic $\frac{(x + 2)^{2}}{7} + (y - 1)^{2} = 14$ is
A) $\sqrt{\frac{7}{8}}$
B) $\sqrt{\frac{6}{17}}$
C) $\frac{\sqrt{3}}{2}$
D) $\sqrt{\frac{6}{7}}$
Explanation
(d)
Given, $\frac{(x + 2)^{2}}{7 \times 14} + \frac{(y - 1)^{2}}{14} = 1$
Here, $a^{2} = 7 \times 14$ and $b^{2} = 14\$
We know, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{14}{7 \times 14}} = \sqrt{\frac{6}{7}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.