Q. For the ellipse $25x^{2} + 9y^{2} - 150x - 90y + 225 = 0$, the eccentricity $e$ is equal to
A) $\frac{2}{5}$
B) $\frac{3}{5}$
C) $\frac{4}{5}$
D) $\frac{1}{5}$
Explanation
(c)
Given equation can be rewritten as
$\frac{(x - 3)^{2}}{9} + \frac{(y - 5)^{2}}{25} = 1\ \ \ \ \ \lbrack b > a\rbrack$
$\therefore\ e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.