Question

The eccentricity of the hyperbola $4x^2-y^2-8x-8y-28=0$ is

• A. $3$
• B. $\sqrt{5}$
• C. $2$
• D. $\sqrt{7}$
• E. $\sqrt{2}$

Answer 1

The correct option is B $\sqrt{5}$

Given equation of hyperbola is
$4x^2-y^2-8x-8y-28=0$
$\Rightarrow 4(x^2-2x)-(y^2+8y)-28=0$
$\Rightarrow 4(x^2-2x+1-)-(y^2+8y+16-16)-28=0$
$\Rightarrow 4(x-1{)}^2-4-(y+4{)}^2+16-28=0$
$\Rightarrow 4(x-1{)}^2-(y+4{)}^2=16$
$\Rightarrow \frac{(x-1{)}^2}{4}-\frac{(y+4{)}^2}{16}=1$
Here, $a^2=4$ and $b^2=16(\because \frac{x^2}{a^2}-\frac{y^2}{b^2}=1)$
Therefore, eccentricity of hyperbola is
$b^2=a^2(e^2-1)$
$\Rightarrow 16=4(e^2-1)$
$\Rightarrow e^2-1=4$
$\Rightarrow e^2=5$
$\Rightarrow e=\sqrt{5}$