Show that $y=\frac{1}{x}$ is a hyperbola

Question

I am trying to write $y=\frac{1}{x}$ in the standard form for a hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ($a>0$, $b>0$) but I can't figure out the last step. My book says to

First, rotate the coordinate system through an angle $\theta$. This means that the coordinates $x$ and $y$ should be replaced by, respectively, $x\cos\theta+y\sin\theta$ and $-x\sin\theta+y\cos\theta$. Choosing an appropriate $\theta$, we can make the coefficient of $xy$ equal to zero.

With this modification was able to rewrite the equation as

$$1=xy=(x\cos\theta+y\sin\theta)(-x\sin\theta+y\cos\theta)=-x^2(\sin\theta\cos\theta)+xy(\cos^2\theta-\sin^2\theta)+y^2(\sin\theta\cos\theta).$$

To get rid of the $xy$ term I set $\theta=\frac{\pi}{4}$ which simplified the equation to $$1=-\frac{x^2}{4}+\frac{y^2}{4}.$$

But the signs on my $x^2$ and $y^2$ terms are not what they should be with real $a,b$. I assume this can be remedied in the next step, but I'm not sure how.

Next we move the origin to $(x_0,y_0)$, i.e., we replace $x$ by $x+x_0$ and $y$ by $y+y_0$. By choosing an appropriate pair $(x_0,y_0)$ we can transform the equation into the canonical form.

Answer 1

The hyperbola $xy=1$ is rotated at an angle of $45^∘=\pi/4$ relative to the $x$-axis. We can use a transformation to rotate our coordinate system by $45^∘$ to find the equation of this hyperbola relative to a coordinate system centered at the current origin but with its positive $x$-axis coincident with the hyperbola’s axis of symmetry. The required transformation is:

\begin{cases} x=\frac{\sqrt 2}{2}x'-\frac{\sqrt 2}{2}y'\\ y=\frac{\sqrt 2}{2}x'+\frac{\sqrt 2}{2}y' \end{cases}

Substituting these into the equation for $x$ and $y$, we get:

$$\left(\frac{\sqrt 2}{2}x'-\frac{\sqrt 2}{2}y'\right)\left(\frac{\sqrt 2}{2}x'+\frac{\sqrt 2}{2}y'\right)=1$$

Simplifying, we get:

$$\frac{x'^2}{2}-\frac{y'^2}{2}=1$$

In this new rotated coordinate system, we see the this hyperbola has a center at $(0,0)$, vertices at $(−\sqrt 2,0)$ and $(\sqrt 2,0)$, foci at $(−2,0)$ and $(2,0)$, and an eccentricity, $e=\sqrt 2$.

follow-up

To obtain the equation of the conic $y=\frac kx$ in this situation, we consider the equation $x^2-y^2=a^2$ of the equilateral hyperbola with asymptotes the bisectors of the quadrants obtained in the previous paragraph and rotate it $45°$ counterclockwise.

The coordinates of the point $ P'$ obtained by making a rotation of point $P$ with center in the origin and of amplitude a counterclockwise are given by: \begin{cases} x'=\frac{\sqrt 2}{2}x+\frac{\sqrt 2}{2}y\\ y'=-\frac{\sqrt 2}{2}x+\frac{\sqrt 2}{2}y \end{cases} Substituting into the equation of the hyperbola gives:

$$\left(\frac{\sqrt 2}{2}x+\frac{\sqrt 2}{2}y\right)^2-\left(-\frac{\sqrt 2}{2}x+\frac{\sqrt 2}{2}y\right)^2=1$$ Developing this gives:

$$xy=\frac{a^2}2$$

and by placing $\frac{a^2}2=k$ we obtain:

$$xy=k \iff y=\frac kx, \quad k>0, x\ne 0$$

Answer 2

You can also reverse engenieering it using the remarkable identity $(A+B)(A-B)=A^2-B^2$

$\displaystyle\frac{X^2}{a^2}-\frac{Y^2}{b^2}=\underbrace{\Big(\frac Xa+\frac Yb\Big)}_x\cdot\underbrace{\Big(\frac Xa-\frac Yb\Big)}_y=1$

Solving for $X,Y$ you get $\begin{cases}X=\frac a2(x+y)\\Y=\frac b2(x-y)\end{cases}$

Now if you want the change of variable to be orthonormal you need $a=b$ and the norm $\sqrt{\frac{a^2+b^2}4}=1$ hence $\sqrt{2}$

Answer 3

We may write:

$$x.y=(\frac x2+\frac y2)^2-(\frac x2-\frac y2)^2=1$$

this manupolation deduces that transformation can be in this form:

$X=x+y$

$Y=x-y$

such that:

$$(\frac X2)^2-(\frac Y2)^2=1$$