Question

The eccentricity of the conic
$ 4x^2 +16 y^2 -24 x -32 y = 1 $ is :

• A. $ \frac{1}{2} $
• B. $ \sqrt {3} $
• C. $ \frac{\sqrt 3}{2} $
• D. $ \frac{\sqrt 3}{4} $

Answer 1

Answer: (C)

We have, $4x^{2} + 16y^{2} -24x -32y = 1 $
$ \Rightarrow 4x^{2} -24 x + 16y^{2} -32y = 1 $
$ \Rightarrow 4\left(x^{2} -6x\right) + 16 \left(y^{2}-2y\right) = 1$
$\Rightarrow 4\left(x^{2} -6x +9 \right) +16 \left(y^{2} -2y +1\right) -36 -16 = 1 $
$ \Rightarrow 4\left(x-3\right)^{2} +16\left(y-1\right)^{2} = 53$
$ \Rightarrow \frac{\left(x-3\right)^{2}}{\frac{53}{4}} + \frac{\left(y-1\right)^{2}}{\frac{53}{16}} = 1$
On compairing with $\frac{x^{2}}{a^{2} } + \frac{y^{2}}{b^{2}} = 1$, we get
$ a^{2} = \frac{53}{4}$ and $b^{2} = \frac{53}{16} $
$\therefore $ Eccentricity of ellipse is $e = \sqrt{\frac{a^{2}-b^{2}}{a^{2}}}$
$\Rightarrow e = \sqrt{\frac{\frac{53}{4}-\frac{53}{16}}{\frac{53}{4}}} $
$ \Rightarrow e = \frac{\sqrt{3}}{2} $