Identify conics and find out center, major axes, foci, eccentrity and if it's a hyperbola, its asymptotes.
a) $xy+x-y=2$
b) $x^2+2xy+y^2=4x-4y+4$
I've rewritten a) as $y= {2-x \over x-1}$ and b) $(x+y)^2=4(x-y+1)$ but havent really gotten anything out of this, I drew them by a program and they were skewed (because of $xy$ being in the term?)
For a)
The equation can be written as $$y=-1+\frac{1}{x-1}$$ This is a hyperbola whose asymptotes are $x=1$ and $y=-1$, and the center is $C(1,-1)$. The major axis is given by $y-(-1)=1\cdot (x-1)$, i.e. $y=x-2$. (The major axis bisects the angle between the asymptotes, so the slope of the major axis is $1$.)
Let $A(2,0),B(1,1)$. ($A$ is the intersection point of the major axis with the hyperbola, and $B$ is the intersection point of the line $x=1$ with the line, passing through $A$, parpendicular to the major axis.) The foci are on the major axis, so we can set $F(a,a-2)$. Having $CF=\sqrt{CA^2+AB^2}$ gives $a=1\pm\sqrt 2$, and the eccentricity is $CF/CA=\sqrt 2$.
For b)
You can rotate it to have the form without $xy$ term.
If we rotate it by $-\frac{\pi}{4}$ around the origin, then we get $$\left(x\cos\left(-\frac{\pi}{4}\right)+y\sin\left(-\frac{\pi}{4}\right)\right)^2$$$$+2\left(x\cos\left(-\frac{\pi}{4}\right)+y\sin\left(-\frac{\pi}{4}\right)\right)\left(-x\sin\left(-\frac{\pi}{4}\right)+y\cos\left(-\frac{\pi}{4}\right)\right)$$$$+\left(-x\sin\left(-\frac{\pi}{4}\right)+y\cos\left(-\frac{\pi}{4}\right)\right)^2$$$$=4\left(x\cos\left(-\frac{\pi}{4}\right)+y\sin\left(-\frac{\pi}{4}\right)\right)-4\left(-x\sin\left(-\frac{\pi}{4}\right)+y\cos\left(-\frac{\pi}{4}\right)\right)+4,$$ i.e. $$y=-\frac{\sqrt 2}{4}x^2+\frac{\sqrt 2}{2}$$ so this is a parabola whose major axis is $x=0$, focus $(0,0)$, eccentricity $1$.
Therefore, rotating back by $+\frac{\pi}{4}$ around the origin gives that