Find the eccentricity of the conic represented by $x^{2}-y^{2}-4 x+4 y+16=0$
Find the eccentricity of the conic represented by $x^{2}-y^{2}-4 x+4 y+16=0$
2
$\sqrt{2}$
$2 \sqrt{2}$
$3 \sqrt{2}$
Solution
We have $x^{2}-y^{2}-4 x+4 y+16=0$
$\Rightarrow\left(x^{2}-4 x\right)-\left(y^{2}-4 y\right)=16$
$\Rightarrow\left(x^{2}-4 x+4\right)-\left(y^{2}-4 y+4\right)=-16$
$\Rightarrow(x-2)^{2}-(y-2)^{2}=-16$
$\Rightarrow \frac{(x-2)^{2}}{4^{2}}-\frac{(y-2)^{2}}{4^{2}}=1$
This is rectangular hyperbola, whose eccentricity is always $\sqrt{2}$.