Question

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

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

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

$4x^2+16y^2-32x-32y=1$
$\Rightarrow 4(x^2-8y+4^2)+16(y^2-2y+1)=1+16+64$
$\Rightarrow 4(x-4{)}^2+16(y-1{)}^2=81$
$\Rightarrow \frac{(x-4{)}^2}{\left(\frac{81}{4}\right)}+\frac{(y-1{)}^2}{\left(\frac{81}{16}\right)}=1$
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,
we get
$a=\frac{9}{2},b=\frac{9}{4}$ clearly $(a>b)$
$\therefore e^2=1-\frac{b^2}{a^2}$
$\Rightarrow e^2=1-\frac{1}{4}$
$\Rightarrow e=\frac{\sqrt{3}}{2}$