How to prove that the graph of function $y=\frac{x}{\sqrt{3}}+\frac{1}{x}$ is a hyperbola?

Question

How to prove that the graph of function $y=\frac{x}{\sqrt{3}}+\frac{1}{x}$ is a hyperbola? Actually,I want to know this question can be proved in polar coordinates with rotation? Thanks a lot

Answer 1

We have

$ y = \dfrac{x}{\sqrt{3}} + \dfrac{1}{x}$

Multiplying through by $\sqrt{3} x$

$ \sqrt{3} x y = x^2 + \sqrt{3} $

Define the position vector $r = [x, y]^T $ then the above equation can be written in matrix-vector format as

$r^T Q r = c $

where

$Q =\begin{bmatrix} 2 && -\sqrt{3} \\ -\sqrt{3} && 0 \end{bmatrix} , \hspace{30pt} c = - 2 \sqrt{3} $

To find the equation of this conic, we diagonalize $Q$ into $Q = R D R^T $, as follows:

Define $\phi = \frac{1}{2} \tan^{-1} \left(\dfrac{2 (-\sqrt{3} ) }{ 2 - 0 }\right) = \frac{1}{2} \tan^{-1}(-\sqrt{3} ) = - \dfrac{\pi}{6} $

Now, the rotation matrix $R$ is given by

$R = \begin{bmatrix} \cos(\phi) && - \sin(\phi) \\ \sin(\phi) && \cos(\phi) \end{bmatrix} = \begin{bmatrix} \dfrac{\sqrt{3}}{2} && \dfrac{1}{2} \\ -\dfrac{1}{2} && \dfrac{\sqrt{3}}{2} \end{bmatrix} $

And the diagonal elements of the diagonal matrix $D$ are given by

$\begin{equation}\begin{split} D_{11} &= Q_{11} \cos^2(\phi) + Q_{22} \sin^2(\phi) + 2 Q_{12} \sin(\phi) \cos(\phi) \\ &= \dfrac{3}{2} + \dfrac{3}{2} = 3 \end{split}\end{equation}$

$\begin{equation}\begin{split} D_{22} &= Q_{11} \sin^2(\phi) + Q_{22} \cos^2(\phi) - 2 Q_{12} \sin(\phi) \cos(\phi) \\ &= \dfrac{1}{2} - \dfrac{3}{2} = -1\end{split}\end{equation} $

To put it in standard format we have to divide both sides by $c$. Doing so, gives

$ r^T R E R^T r = 1 $

where $E$ is a diagonal matrix defined as $E = \dfrac{D}{c} $. Hence,

$E_{11} = - \dfrac{\sqrt{3}}{2} , E_{22} = \dfrac{1}{ 2 \sqrt{3}} $

Finally, by defining the rotated axes, $w = R^T r $ we get

$ w^T E w = 1 $

And the equation of this conic is

$ - \dfrac{\sqrt{3}}{2} w_1^2 + \dfrac{1}{2 \sqrt{3}} w_2^2 = 1 $

And this an equation of a hyperbola. The major axis (where the hyperbola opens up) is along the direction making $ -\dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{\pi}{3} = 60 ^\circ$ with the horizontal $x$ axis.

This is how the hyperbola looks like in the $xy$ plane