Question

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. $\frac{\sqrt{3}}{4}$

Answer 1

Answer: (C)

We have, $4x^2+16y^2-24x-32y=1$
$\Rightarrow 4x^2-24x+16y^2-32y=1$
$\Rightarrow 4\left(x^2-6x\right)+16\left(y^2-2y\right)=1$
$\Rightarrow 4\left(x^2-6x+9\right)+16\left(y^2-2y+1\right)-36-16=1$
$\Rightarrow 4{\left(x-3\right)}^2+16{\left(y-1\right)}^2=53$
$\Rightarrow \frac{{\left(x-3\right)}^2}{\frac{53}{4}}+\frac{{\left(y-1\right)}^2}{\frac{53}{16}}=1$
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 $e=\sqrt{\frac{a^2-b^2}{a^2}}$
$\Rightarrow e=\sqrt{\frac{\frac{53}{4}-\frac{53}{16}}{\frac{53}{4}}}$
$\Rightarrow e=\frac{\sqrt{3}}{2}$