# Ask Uncle Colin: A rotated hyperbola

Dear Uncle Colin,

I have the curve $y = \frac{1}{x}$ and need to rotate it by 45 degrees clockwise. I can’t figure out what curve I get! Any clues?

Where Asymptotes Lie, Like I Said

Hi, WALLIS, and thanks for your message!

I can think of several ways to do this. Let’s look at some!

### Polar coordinates

One way is to convert it into polar coordinates, adjust the angle, and convert back.

If $y = \frac{1}{x}$, then $xy = 1$; since $x = r\cos(\theta)$ and $y = r\sin(\theta)$, that gives $r^2 \sin(\theta)\cos(\theta) = 1$.

Better yet, we can make that $r^2 \sin(2\theta) = 2$.

Rotating it by $-\piby 4$ radians ((as Euler intended)) turns $\theta$ into $\theta+\piby 4$, so we have - for the rotated curve - $r^2 \sin\left(2\theta + \piby 2\right)=2$.

Now, $\sin(2\theta + \piby2) \equiv -\cos(2\theta)$, so our curve is $-r^2 \cos(2\theta) = 2$

Let’s get that back into the $x-y$ plane. $\cos(2\theta) \equiv 2\cos^2(\theta) - 1$, and $\cos(\theta) = \frac{x}{r}$, while $r^2 \equiv x^2 + y^2$ so the curve is $-(x^2+y^2)\br{2\frac{x^2}{x^2+y^2}- 1} = 2$.

Rearranging: $-\br{x^2 + y^2}\br{2x^2 - (x^2 + y^2)} = 2\br{x^2 + y^2}$, or $\br{x^2+y^2}\br{y^2-x^2} = 2\br{x^2 + y^2}$.

We also know that $x^2 + y^2 > 0$ because $x\ne0$, so we can divide by it to get $y^2 = x^2 - 2$.

Does that make sense? As $x$ gets large, $y$ approaches $\pm x$, as we would like; the curve also passes through $\br{\sqrt{2},0}$, which is where we want it to be.

### Matrices

If we represent $y = \frac{1}{x}$ as $\colvectwo{t}{\frac{1}{t}}$ for $t \ne 0$ and rotation by $-\piby4$ as $\frac{1}{\sqrt{2}}\mattwotwo{1}{1}{-1}{1}$, then the transformed curve is $\frac{1}{\sqrt{2}}\colvectwo{t + \frac{1}{t}}{\frac{1}{t}-t}$.

So, in parametric form, we have $y = \frac{1}{\sqrt{2}} \br{\frac{t^2+1}{t}}$ and $x = \frac{1}{\sqrt{2}}\br{\frac{1-t^2}{t}}$

Then we need to isolate $t$, which is a bit messy.

- $xt\sqrt{2} = 1 + t^2$
- $yt\sqrt{2} = 1 - t^2$

If we square both of those:

- $2 x^2 t^2 = 1 + 2t^2 + t^4$
- $2 y^2 t^2 = 1 - 2t^2 + t^4$

Subtract:

- $2t^2 \br{x^2 - y^2} = 4t^2$

Since $t^2 \ne 0$, this reduces to $x^2 - y^2 = 2$, as before.

### Whoosh

“Should have known I’d find you here, sensei.”

A better substitution is $y = e^t$ and $x = e^{-t}$. Then your bullet points are $xe^t \sqrt{2} = 1 + e^{2t}$ and $ye^t\sqrt{2} = 1 - e^{2t}$.

“And that’s better? You’re right, your bullets are - as always - better than my bullets.”

Alternatively, $x = \sqrt{2}\cosh(x)$ and $y = \sqrt{2}\sinh(x)$. Thus $x^2 - y^2 = 2$.

### Whoosh

Hope that helps!

- Uncle Colin