Solving Two-Variable Problems: F(x,y)=-(y/x)^2+h(xy)

[Bleys]

There are some problems that ask you to find an equation with some information and I'm having trouble with it
For example, z= F(x,y) = -(y/x)^2 + h(xy), for some arbitrary function h.
They give you the condition that F(1,y) = y^2, and to find z now.
So with obvious substitution you get h(y) = 0
But how does that tell you anything else about z?
Does it mean h(t) = (t - y) ? Aren't there an infinite amount of ways to express h anyway (like ln(t-y+1) for example)? I doubt they are asking you for an explicit form, since no other information is available, but how do you go about solving a problem like this. I'm having a little problems understanding what exactly it's saying.
Does h(y)=0 mean y is a solution to the function h when the domain is restricted to points (1,y)?

 

[djeitnstine]

Looking carefully at your function when you insert x=1 you have $$F(1,y)=-y^2 + H(1,y)$$ if the negative sign is in the right place. The answer should be obvious what $$H(x,y)$$ is

 

[HallsofIvy]

There are some problems that ask you to find an equation with some information and I'm having trouble with it
For example, z= F(x,y) = -(y/x)^2 + h(xy), for some arbitrary function h.
They give you the condition that F(1,y) = y^2, and to find z now.
So with obvious substitution you get h(y) = 0
But how does that tell you anything else about z?
Does it mean h(t) = (t - y) ? Aren't there an infinite amount of ways to express h anyway (like ln(t-y+1) for example)? I doubt they are asking you for an explicit form, since no other information is available, but how do you go about solving a problem like this. I'm having a little problems understanding what exactly it's saying.
Does h(y)=0 mean y is a solution to the function h when the domain is restricted to points (1,y)?

h is a function of a single variable- its domain is a subset of the set of all real numbers, not points in the plane.

F(1,y)= -(y/1)^2+ h(1y)= -y^2+ h(y)= y^2, but that does NOT give you "h(y)= 0" it gives "h(y)= 2y^2". F(x,y)= -(y/x)^2+ 2(xy)^2 for all x and y.

Did you mean F(x,y)= (y/x)^2+ h(xy) and F(1,y)= y^2? In that case F(1,y)= (y/1)^2+ h(1y)= y^2 so h(y)= 0 for all y and so F(x,y)= (y/x)^2 for all x and y.

Did you mean F(x,y)= -(y/x)^2+ h(xy) and F(1,y)= -y^2? In that case F(1,y)= -(y/1)+ h(1y)= -y^2 so again h(y)= 0. F(x,y)= -(y/x) for all x and y.