Unit Hyperbola

Consider the following equation:

$$x^2-y^2=1$$

Note

Notice how the equation compares to the equation of a unit circle, $x^2+y^2=1$.

Spoiler alert: This is the equation of the unit hyperbola.

We can rearrange this to make $y$ the subject:

$$y^2=x^2-1$$ $$y=\pm \sqrt{x^2-1}$$

Consider how the function would behave as we make the value of $x$ very big in either the positive or negative direction. The $-1$ alongside it under the square root becomes near-insignificant, leaving the square and square root to cancel out:

$$\begin{array}{c}y\approx \pm \sqrt{(\text{really big }x{)}^2}\\ y\approx \pm \text{really big }x\end{array}$$

We know that as $x$ approaches $\pm \infty$, the function approaches $y=\pm x$, meaning there are asymptotes of $y=x$ and $y=-x$. We also know that the function is symmetric across the x-axis (because of the $\pm$).

Now consider what has to happen for $y$ to be zero, to figure out where this function would cross the x-axis:

$$\begin{array}{c}0=\pm \sqrt{x^2-1}\\ x^2-1=0\\ x^2=1\\ x=\pm 1\end{array}$$

We now know that the function must also pass through the points $(1,0)$ and $(-1,0)$ as well.

This is enough to provide a basic intuition for the shape of the graphed function:

Let’s finally take a look at the function in all its hyperbolic glory:

Yippee!