Q. The eccentricity of the conic $4x^{2} + 16y^{2} - 24x - 32y = 1$ is

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

(c)
Given equation can be rewritten as
$\Rightarrow 4\left( x^{2} - 6x + 9 \right) + 16\left( y^{2} - 2y + 1 \right) - 36 - 6 = 1$
$\Rightarrow \ \frac{(x - 3)^{2}}{\frac{53}{4}} + \frac{(y - 1)^{2}}{\frac{53}{16}} = 1$
Here , $a^{2}\ = \frac{53}{4}\ \$and$\$ $b^{2} = \frac{53}{16}$
$\therefore$ Eccentricity of ellipse is $e = \frac{\sqrt{a^{2} - b^{2}}}{a^{2}}$
$\Rightarrow \ e = \frac{\sqrt{\frac{53}{4} - \frac{53}{16}}}{\frac{53}{4}}$
$\Rightarrow \ e = \frac{\sqrt{3}}{2}$

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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