$S_f(r)=\{x\in A: f(x)=r\}$ is a plane curve

Question

There is the following statement in our book:

Let $r\in\mathbb{R}$ be a constant and $f:A \rightarrow \mathbb{R}, A\subset \mathbb{R}^2$ a function.

Now $S_f(r)=\{x\in A: f(x)=r\}$ is usually a plane curve.

I am having a bit of a trouble understanding what this means and looks like.

I also don't know how to use it to find the plane curves of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}, f(x_1,x_2)=2x_1^2-x_2^2$.

Answer 1

$S_f(r)=\{X\in A: f(X)=r\}$ can be considered as the intersection $$\{(x,y,f(x,y))|(x,y)\in A\}\cap\{(x,y,z)|z=r\}.$$ The first one is a surface and the second is plane, both are in $\mathbb{R^3}$.

Answer 2

The background is the implicit function theorem. If $f(x,y)$ is a continuous differentiable map from $\mathbb R^2$ into $\mathbb R$ and $\dfrac{\partial f}{\partial y} \neq 0$, you can find (locally) a map $g$ such that $$f(x,y)=r \Leftrightarrow y=g(x)$$

Which represents a curve.

You can apply that to the example you gave at the end of your question.