Question

How to show that surface f(x,y)=(x^2+2y^2, 66x^2-24xy+59y^2) is tangent to xy plane at the origin?

Answer 1

I am not sure whether my logic is valid or not. I just give out my opinion :)
The equation of xy plane is z =0, If we can show the tangent plane of the surface is z =0 at (0,0) then we are done.
we have
\[f_x(x,y) = x^2+2y^2\\f_y(x,y) = 66x^2-24xy+59y^2\]
At (0,0,0) we have \(f_x(0,0) =0\) and \(f_y(0,0) =0\)
that show z -0 = 0(x-0)+0(y-0) =0 or z =0
so at (0,0,0) the tangent plane is z =0 or xy plane

Answer 2

@OOOPS you're right about the approach, but \(f(x,y)\) is a function \(f:\mathbb{R}^2\to\mathbb{R}^2\) (two inputs, two outputs), not \(f:\mathbb{R}^2\to\mathbb{R}\). There's a bit more work involved, I think.

Answer 3

Here's an example worked out: http://www.physicsforums.com/showthread.php?t=491376