Q. The eccentricity of the conic $36x^{2} + 144y^{2} - 36x - 96y - 119 = 0$ is

A) $\frac{\sqrt{3}}{2}$
B) $\frac{1}{2}$
C) $\frac{\sqrt{3}}{4}$
D) $\frac{1}{\sqrt{3}}$
Explanation

(a)
Given equation can be rewritten as
$36\ \left( x^{2} - x + \frac{1}{4} \right) + 144\ \left( y^{2} - \frac{2}{3}y + \frac{1}{9} \right) = 144$
$\Rightarrow \frac{\left( x - \frac{1}{2} \right)^{2}}{4} + \frac{\left( y - \frac{1}{3} \right)^{2}}{1} = 1$
$\therefore\ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

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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