I was doing a sample exam for my final tommorow and I did not understand how to solve the following question. Maybe it's relevant to note that during this math course we have been talking about the Implicit Function Theorem and Inverse Mapping Theorem.
We were supposed to sketch the equation g(x,y), but I have no clue how they managed to come to the conclusion of that in the picture. For example, how do they know its a hyperbola and how did they find those asymptotes in a multi-variate equation? I would greatly appreciate any input.
Thank you very much in advance.
Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = \{(x,y) \in \mathbb R^2 : g(x,y) = 0\}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = \{(x,y) \in \mathbb R^2 : z = 0\},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.
Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$\begin{align}x^2 - y^2 + 2y - 2 & = 0 \\ x^2 -1(y^2 - 2y) - 2 &= 0 \tag{Factor out $-1$ from $y$'s}\\ x^2 - (\underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 \tag{Complete square}\\ x^2 - [(y-1)^2 - 1] - 2 &= 0\\ x^2 - (y-1)^2 + 1 - 2 &= 0 \tag{Simplify}\\ x^2 - (y-1)^2 - 1 &= 0 \tag{Simplify}\\ (x - 0)^2 - (y - 1)^2 = 1. \tag{Move $-1$ over}\end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.