y = \sqrt{x^2-1}

The Upper Half of a Hyperbola $y = \sqrt{x^2-1}$

$y = \sqrt{x^2 - 1}$ is the upper half of the rectangular hyperbola $x^2 - y^2 = 1$ ¹. Where the existing $y = \sqrt{x^2+1}$ is a single connected curve, changing one sign inside the radical splits this one into two separate branches.

Domain and range

The domain is $x \leq -1$ or $x \geq 1$
The range is $y \geq 0$
Even function
There is no curve at all on $-1 < x < 1$

$x$	$y$
$\pm 1$	$0$
$\pm\sqrt{2}$	$1$
$\pm 2$	$\sqrt{3}$

The elements of the hyperbola

Together with the lower half, the full curve is symmetric about the $x$ -axis and the origin as well.

Element	Value
Vertices	$(\pm 1, 0)$
Foci	$(\pm\sqrt{2}, 0)$
Eccentricity	$\sqrt{2}$
Asymptotes	$y = \pm x$

Asymptotes

Since $\sqrt{x^2-1} - x = \dfrac{-1}{\sqrt{x^2-1} + x}$ tends to $0$ as $x \to +\infty$ , the line $y = x$ is an asymptote, and on the left branch the asymptote is $y = -x$ . These two lines are perpendicular, which is why the curve is called a rectangular hyperbola.

Tangents and concavity

The derivative is $y' = \dfrac{x}{\sqrt{x^2-1}}$ , which diverges to $+\infty$ as $x \to 1^{+}$ , so the tangent at the vertex is vertical. The second derivative $y'' = -\dfrac{1}{(x^2-1)^{3/2}}$ is always negative, so both branches are concave down.

Curve	Domain	Concavity
$y = \sqrt{x^2+1}$	all real numbers	concave up
$y = \sqrt{x^2-1}$	$\|x\| \geq 1$	concave down

The two curves share their asymptotes but bend in opposite directions.

Inverse function

Solving for $x$ on the branch $x \geq 1$ gives $x = \sqrt{y^2+1}$ . That branch and the part of $y = \sqrt{x^2+1}$ with $x \geq 0$ are therefore inverses of one another, and the two graphs are reflections of each other in the line $y = x$ .

Relation to the hyperbolic functions

Setting $x = \cosh t$ and $y = \sinh t$ gives $\cosh^2 t - \sinh^2 t = 1$ , so the point always lies on the hyperbola. The right branch is exactly the locus traced as $t$ runs over all the reals, with the upper half corresponding to $t \geq 0$ . This parametrisation is precisely why these are called hyperbolic functions.

Moreover the hyperbolic sector determined by the origin, $(1, 0)$ and $(\cosh t, \sinh t)$ has area exactly $\dfrac{t}{2}$ , matching the way the angle on the unit circle is twice the area of the corresponding circular sector.

Applications

A hyperbola is the locus of points whose distances to two fixed points differ by a constant.

Hyperbolic navigation, fixing a position from the difference in arrival times of two signals²
The worldline of an object under constant proper acceleration
Revolving the curve about the $x$ -axis gives the hyperboloid of two sheets

Hyperbola, Wikipedia
Hyperbolic navigation, Wikipedia

The Upper Half of a Hyperbola y=x2−1y = \sqrt{x^2-1}y=x2−1​