Q. The eccentricity of the ellipse $25x^{2} + 16y^{2} - 150x - 175 = 0$ is
A) $\frac{2}{5}$
B) $\frac{2}{3}$
C) $\frac{4}{5}$
D) $\frac{3}{5}$
Explanation
(d)
Given equation of ellipse can be rewritten as
$\frac{(x - 3)^{2}}{16} + \frac{y^{2}}{25} = 0$
The major axis of ellipse is a line parallel to $y$-axis therefore eccentricity of ellipse is given by
$e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.