Consider a branch of the hyperbola  x2−2y2−22x−42y−6=0 with vertex at the point A.let B be one of the end points of its latus rectum. If C is thefocus of the hyperbola near A, then area of the ∆ ABC is:

Equation of the hyperbola is (x−2)24−(y+2)22=1Which can be written as X24−Y22=1 where x=X+2,y=Y−2. Eccentricity of the hyperbola is 4+24=32     Area of the ΔABC=12(AC)(BC)=12(ae−a)×b2a where a2=4,b2=2=12232−2×22=32−1