If f(x+2y, x-2y)=xy, then f(x, y) equals
Let x + 2y = p x – 2y = q Solving we got x=p+q2 y=p−q2∴f(p,q)=p2−q28F(x,y)=x2−y28
Let f(x,y)=√x2+y2+√x2+y2−2x+1+√x2+y2−2y+1+√x2+y2−6x−8y+25∀x,yϵR, then