We have,
`x^(2)-y^(2)-4x+4y+16=0`
`implies(x^(2)-4x)-(y^(2)-4y)=16`
`implies(x^(2)-4x+4)-(y^(2)-4y+4)=-16`
`implies(x-2)^(2)-(y-2)^(2)=-16`
`implies((x-2)^(2))/(4^(2))-((y-2)^(2))/(4^(2))=1`
Shifting the origin at `(2,2)` we obtain
`(X^(2))/(4^(2))-(Y^(2))/(4^(2))=-1`, where `x=X+2`, `y=Y+2`
Clearly , it is a rectangular hyperbola, whose eccentricity is always `sqrt(2)`.
ALITER In the given equation, we have
`"Coeff. of" x^(2)+"Coeff. of" y^(2)=0`
So, the hyperbola is a rectangular hyperbola.
Hence, eccentricity `=sqrt(2)`.
