Q. The curve represented by the equation $4x^{2} + 16y^{2} - 24x - 32y - 12 = 0$ is
A) A parabola
B) A pair of straight lines
C) An ellipse with eccentricity $\frac{1}{2}$
D) An ellipse with eccentricity $\frac{\sqrt{3}}{2}$
Explanation
(d)
The given equation can be rewritten as
$4x^{2} - 24x + 36 + 16y^{2} - 32y + 16 - 36 - 16 - 12 = 0$
$\Rightarrow {(2x - 6)}^{2}{+ (4y - 4)}^{2} = 64$
$\Rightarrow \frac{{(x - 3)}^{2}}{16} + \frac{{(y - 1)}^{2}}{4} = 1$
The represents an ellipse and $a^{2} = 16$, $b^{2} = 4$
$\therefore e = \sqrt{1 - \frac{4}{16}} = \frac{\sqrt{3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.