Back in school we were taught that the equation for hyperbola is: $$y=\frac1x \tag1$$ The equation used in the university is different: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag2$$
I can't understand how they are the same and how to get one from another.
What is the relation between these two forms?
On
Those are both hyperbolas, though the second equation is more "general" in the sense that it describes many different hyperbolas (precisely those with their foci on the x-axis). Actually the equation $y = \frac{1}{x}$ describes the graphic of the "homographic function" which is an hyperbola such that it is tangent to both axises. This geometric transformation is necessary to make it an actual fuction.
To obtain the first from the second, first choose $a=b = \sqrt{2}$ in the second equation (in this way you obtain a rectangular hyperbola with perpendicular tangents, then rotate your hyperbola by -45° degrees, which is apply the following transformation: \begin{align} x &= \frac{\zeta +\eta}{\sqrt{2}}\\ y &= \frac{-\zeta +\eta}{\sqrt{2}} \end{align} Thus the second equation becomes \begin{align} \frac{2\zeta\eta}{2} = 1 \end{align} Solving for $\eta$ you obtain \begin{align} \eta = \frac{1}{\zeta} \end{align} Now changing name of your variables you get the first equation. Can you see why now this is a function?
$$xy=1~~~~(1)$$ is rectangular hyperbola but $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~~~~(2)$$ is more standard hyperbola.
Interestingly, (2) can also be written as $$XY=1, X=\frac{x}{a}-\frac{y}{b}, Y=\frac{x}{a}+\frac{y}{b}~~~(3)$$ in the transformed co-ordinate system.
EDIT: We can write $$\begin{bmatrix} X \\ Y \end{bmatrix}=(ab)^{-1} \begin{bmatrix} b & -a \\ b & a \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ This transformation matrix can be belated to the rotataion matrix: $$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos\theta\end{bmatrix}.$$ in a special case.