How can the equation of a hyperbola be $xy=1$?

Question

We know that the standard form of a hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which is not coherent with $xy=1$. So, how can we say that $xy=1$ is a hyperbola?

Answer 1

Actually equations in the form of $Ax^2+2Bxy+C^2+Dx+Ey=F$ can be simplified to an equation with variables $u,v$ depends on $x,y$ such that $\{u,v\}$ is an orthogonal basis for $\Bbb R^2$, i.e. one can rotate the graph in $uv$ coordinate into $xy$ plane.

Let $Ax^2+2Bxy+C^2+Dx+Ey=F$. Then this equation can be written as $$q(x,y)=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}A&B\\B&C\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}D&E\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=F$$ Find the eigenvalues and eigenvectors of $\begin{pmatrix}A&B\\B&C\end{pmatrix}$,namely $v_1,\lambda_1,v_2,\lambda_2$ . Then a rotation matrix$R$ can be defined as $$R=\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}\text{ if }\det\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}=1$$ $$R=\begin{pmatrix}\frac{v_1}{|v_1|}&-\frac{v_2}{|v_2|}\end{pmatrix}\text{ if }\det\begin{pmatrix}\frac{v_1}{|v_1|}&\frac{v_2}{|v_2|}\end{pmatrix}=-1$$ In such way $\begin{pmatrix}x\\y\end{pmatrix}=R\begin{pmatrix}u\\v\end{pmatrix}$. One can verify that the $xy$ term in the original given equation would be eliminated. Draw the graph with variables $u,v$ on the $uv$ plane. Then draw $xy$ plane with $u,v$ axis on it, then just draw the graph with respect to $u,v$ axis.

For practice, try graph $x^2+2xy+y^2-3x+y=0$, which is a parabola.

Answer 2

They are both hyperbolas, but they are rotated $45^\circ$ compared to each other. Specifically, $xy=1$ is congruent to $\frac{x^2}2-\frac{y^2}2=1$.

Answer 3

(Adding to @Arthur)

Consider the rectangular hyperbola $x^2 - y^2 = 2$.

Let's take a point (a, b) of the hyperbola and rotate it by π/4 radian as follows to get the point (c, d).

In polar coordinates, $(a, b)= (r\cosθ, r\sinθ)$

$(c, d) = (r \cos(θ+\fracπ4), r \sin(θ + \fracπ4)) = (\frac{a - b}{\sqrt2}, \frac{a + b}{\sqrt2})$

Writing a and b in terms of c and d gives

$(a, b) = (\frac{c + d}{\sqrt2}, \frac{d - c}{\sqrt2})$

Since (a, b) lies on the hyperbola,

$a^2 - b^2 = 2\\ \dfrac{(c + d)^2}2 - \dfrac{(d - c)^2}2 = 2\\ cd = 1$

So the equation of the new hyperbola is:

$xy=1$