Question

The eccentricity of the conic represented by $x^2-y^2-4x+4y+16=0$ is :

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

Answer 1

The correct option is B $\sqrt{2}$

$x^2-y^2-4x+4y+16=0$
$\Rightarrow (x^2-4x)-(y^2-4y)=-16$
$\Rightarrow (x^2-4x+4)-(y^2-4y+4)=-16$
$\Rightarrow {(x-2)}^2-{(y-2)}^2=-16$
$\Rightarrow -\frac{{(x-2)}^2}{4^2}+\frac{{(y-2)}^2}{4^2}=1$
Shifting the origin at $(2,2)$, we obtain
$-\frac{X^2}{4}+\frac{Y^2}{4}=-1$ where $x=X+2,y=Y+2$.
Clearly it is a rectangular hyperbola whose eccentricity is always $\sqrt{2}$
Hence, option 'B' is correct.