Q. For the ellipse $24x^{2} + 9y^{2} - 120x - 90y + 225 = 0$, the eccentricity is equal to

A) $\frac{2}{5}$
B) $\frac{3}{5}$
C) $\sqrt{\frac{15}{24}}$
D) $\frac{1}{5}$
Explanation

(c)
Given equation can be written as
$24\left( x^{2} - 5x + \frac{25}{4} \right) + 9\left( y^{2} - 10y + 25 \right)$
$+ 225 - 150 - 225 = 0$
$\Rightarrow \ \ \frac{\left( x - \frac{5}{2} \right)^{2}}{\frac{150}{24}} + \frac{(y - 5)^{2}}{\frac{150}{9}} = 1$
$\therefore\ e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{9}{24}} = \sqrt{\frac{15}{24}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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