Q: Give an example for a function $f(x,y)$ continious partial derivatives that satisfies that following:
- $z=f(x,y)$ is not a plane
- $z=f(x,y)$ passes through the point $(6,4,3)$
- The tangent plane to the surface $z=f(x,y)$ at the point $(6,4,3)$ is $$ -4(x-6) +1(y-4) +z -3 = 0 \\~\\ $$
My take: By looking at the tangent plane equation we can tell that: $$ G(x) = f(x,y) -z \\ ~\\ \nabla G(6,4,3) = (-4,1,1) $$ And I don't know how to procceed any further than that. I can think about countless random options, but what is the way to tackle this problem?
From algebraic geometry I'm used to the tangent cone strategy for finding the tangent plane (or line) of a variety given implicitly by one (or, say, two) polynomial equation(s). In its simplest form this method amounts to writing a polynomial around a point (or translating to the origin) and investigating the polynomial there. If your point is on the variety (read surface) the constant term vanishes. The zero set of the linear terms give the tangent plane, etc. The flip side of this is that if you have the tangent plane you can add higher order terms around the point to get surfaces that have this plane as tangent plane. And if you don't add terms of higher order in $(z-3)$ it will still be expressible in polynomials as $z=f(x,y).$
One of the simpler ways of answering your question is then
$$(x-6)^2+(y-4)^2-4(x-6) +1(y-4) +1(z-3) = 0.$$