What is the key difference between a hyperbola and an ellipse equation?

A hyperbola uses a minus sign (-) between the terms, while an ellipse uses a plus sign (+). The minus sign causes the hyperbola to split into two separate branches opening away from each other, whereas the plus sign creates a single closed curve. This single operator difference fundamentally changes the shape and behavior of the entire curve.

How do I find the asymptotes of a hyperbola from its equation?

For the hyperbola \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, the asymptotes pass through the center (h, k) with slopes \pm \frac{b}{a}. Write the asymptote equations using Point-Slope Form: y - k = \pm \frac{b}{a}(x - h). The asymptotes guide the shape of the branches as they extend toward infinity.

What do a and b represent, and how do they differ from an ellipse?

In this hyperbola form, a is the semi-transverse axis length—the distance from the center to each vertex along the horizontal direction. The value b is the semi-conjugate axis length, which does not correspond to any point on the hyperbola itself, but instead controls the steepness of the asymptotes via the slope \frac{b}{a}. In an ellipse, both a and b represent points on the curve; in a hyperbola, only a does.

Can I shift or rotate a hyperbola like other curves?

Yes, the (h, k) in this equation shifts the center away from the origin, and this form represents a hyperbola with axes parallel to the coordinate axes. Rotated hyperbolas exist but require a more complex equation that includes an xy term, which is beyond the scope of this standard form.

Equation of a Hyperbola

The equation of a hyperbola describes all points forming a hyperbola with center at (h, k), used when modeling two-branch curves in coordinate geometry.

\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1

Solve a problem with Equation of a Hyperbola

Type the problem. The solver will use Equation of a Hyperbola where Equation of a Hyperbola is the right tool, and tell you when it is not.

What each symbol means

What Equation of a Hyperbola takes

Symbol	Meaning
$a$	The semi-transverse axis length; the horizontal distance from the center $(h,k)$ to each vertex; students often confuse it with the denominator $a^2$ , causing wrong vertex positions.
$b$	The semi-conjugate axis length; it is not itself a point on the curve but determines the asymptote slopes $\pm \frac{b}{a}$ .
$h$	The x-coordinate of the center; leaving it out shifts the hyperbola horizontally to the origin.
$k$	The y-coordinate of the center; omitting it shifts the hyperbola vertically to the x-axis.
$x$	The x-coordinate of any point on the hyperbola; it ranges over all real numbers except the interval $[h-a, h+a]$ for this horizontal-opening form.
$y$	The y-coordinate of any point on the hyperbola; not all $y$ values are possible, only those that satisfy the equation for a chosen $x$ .

When to use it

Use this when you need to find the coordinates of all points that form a hyperbola or when you're given a hyperbola and need to identify its key features like center and axis lengths.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Check if a point lies on a hyperbola

Problem

Determine whether the point

(5, 3)

lies on the hyperbola

\frac{(x-1)^2}{9} - \frac{(y-2)^2}{4} = 1

$\frac{(5-1)^2}{9} - \frac{(3-2)^2}{4}$
Substitute $x = 5$ and $y = 3$ into the hyperbola equation.
$\frac{16}{9} - \frac{1}{4}$
Simplify the numerators: $(5-1)^2 = 16$ and $(3-2)^2 = 1$ .
$\frac{64}{36} - \frac{9}{36} = \frac{55}{36}$
Convert to a common denominator of 36: $\frac{16}{9} = \frac{64}{36}$ and $\frac{1}{4} = \frac{9}{36}$ .
$\frac{55}{36} \neq 1$
The result is not equal to 1, so the point does not satisfy the equation.

Answer: $(5, 3) \text{ does not lie on the hyperbola}$

We verified membership by substitution: a point is on the hyperbola if and only if its coordinates satisfy the equation exactly. Since our result was $\frac{55}{36}$ instead of 1, the point is not on the curve.

2. Find the missing parameter of a hyperbola given a point

Problem

A hyperbola has the form

\frac{(x-1)^2}{4} - \frac{(y-2)^2}{b^2} = 1

. If the point

(7, 6)

lies on the hyperbola, find

b^2

$\frac{(7-1)^2}{4} - \frac{(6-2)^2}{b^2} = 1$
Substitute $x = 7$ and $y = 6$ into the equation.
$\frac{36}{4} - \frac{16}{b^2} = 1$
Calculate the squares: $(7-1)^2 = 36$ and $(6-2)^2 = 16$ .
$9 - \frac{16}{b^2} = 1$
Simplify the first fraction: $\frac{36}{4} = 9$ .
$\frac{16}{b^2} = 8$
Rearrange by subtracting 9 from both sides; note that $1 - 9 = -8$ , so $-\frac{16}{b^2} = -8$ .
$b^2 = 2$
Solve for $b^2$ by division: $b^2 = \frac{16}{8} = 2$ .

Answer: $b^2 = 2$

We substituted the given point into the equation with unknown $b^2$ , then solved algebraically. This technique works whenever one parameter is unknown but other points on the hyperbola are known.

3. Find a hyperbola equation from a cyclist's training route

Problem

A cyclist trains on a hillside where the path follows a hyperbola centered at a base camp at

(0, 0)

. The cyclist's outbound and return trails are symmetric, reaching a minimum horizontal distance of 3 km from camp. The cyclist passes through the point 5 km east and 4 km north of camp. Find the equation of the hyperbola modeling the training route.

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Since the center is at the origin $(0, 0)$ and the branches open horizontally, use this standard form.
$a = 3, \text{ so } a^2 = 9$
The minimum horizontal distance from camp to the trail is 3 km, which is the semi-transverse axis length $a$ .
$\frac{5^2}{9} - \frac{4^2}{b^2} = 1$
The cyclist passes through $(5, 4)$ , so substitute $x = 5$ and $y = 4$ .
$\frac{25}{9} - \frac{16}{b^2} = 1$
Evaluate the squares: $5^2 = 25$ and $4^2 = 16$ .
$\frac{25}{9} - 1 = \frac{16}{b^2}$
Isolate $\frac{16}{b^2}$ by subtracting 1 from both sides.
$\frac{16}{9} = \frac{16}{b^2}$
Simplify the left side: $\frac{25}{9} - \frac{9}{9} = \frac{16}{9}$ .
$b^2 = 9$
Since both fractions have numerator 16, equate the denominators: $b^2 = 9$ .

Answer: $\frac{x^2}{9} - \frac{y^2}{9} = 1$

We used the constraint that the trail has minimum horizontal extent 3 km to set $a = 3$ , then leveraged the point $(5, 4)$ to solve for $b^2$ . This combined approach—using geometric constraints and coordinate points—is essential for constructing hyperbola equations from real-world scenarios.

Common mistakes

Where Equation of a Hyperbola usually goes wrong

The mistake	Why it is wrong	The fix
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ or $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$	The minus sign between the terms is essential to a hyperbola; using a plus sign gives an ellipse, and switching the order changes which direction the hyperbola opens.	Always verify the middle operator is a minus and the $(x-h)^2$ term comes first in the standard form.
Reading $a = 9$ directly from the denominator in $\frac{(x-h)^2}{9}$ instead of taking the square root to get $a = \sqrt{9} = 3$ .	The value $a$ is the semi-transverse axis length, not the denominator itself; $a^2$ is the denominator.	Always take the square root of the denominator under the positive term: if $a^2 = 9$ , then $a = 3$ .
Using $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ (the vertical form) when the problem gives a horizontal-opening hyperbola, or vice versa.	These two forms open in perpendicular directions; using the wrong form yields incorrect vertices and asymptotes.	Identify which variable term is positive: if $(x-h)^2$ is positive, the hyperbola opens horizontally; if $(y-k)^2$ is positive, it opens vertically.

Tips and when to use something else

The value $a$ always appears in the denominator under the positive (first) term; it governs the distance from the center to the vertices along the axis of symmetry.
The asymptotes of this hyperbola have slopes $\pm \frac{b}{a}$ and pass through the center $(h, k)$ —use Point-Slope Form to write their equations when needed.
A hyperbola opens outward (away from the center) into two branches, unlike an ellipse which forms a single closed curve; this behavior is why the equation has a minus sign.
To classify a conic from the general form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ as a hyperbola, check the discriminant: if $AC < 0$ , it is a hyperbola.

Related formulas

Equation of an EllipseEquation of an Ellipse describes an oval-shaped curve given its center, horizontal stretch, and vertical stretch; use it to model or graph any elliptical shape.Translating a GraphTranslate graphs using y = f(x - h) + k to shift any function horizontally by h and vertically by k units, needed when repositioning known curves.Point-Slope FormPoint-Slope Form lets you write a line equation when you know the slope and one point; use it to quickly write equations or convert to standard form.Distance FormulaCalculates the shortest distance between two points on a coordinate plane using their coordinates and the Pythagorean theorem.

Frequently asked questions

What is the key difference between a hyperbola and an ellipse equation?: A hyperbola uses a minus sign ( $-$ ) between the terms, while an ellipse uses a plus sign ( $+$ ). The minus sign causes the hyperbola to split into two separate branches opening away from each other, whereas the plus sign creates a single closed curve. This single operator difference fundamentally changes the shape and behavior of the entire curve.
How do I find the asymptotes of a hyperbola from its equation?: For the hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ , the asymptotes pass through the center $(h, k)$ with slopes $\pm \frac{b}{a}$ . Write the asymptote equations using Point-Slope Form: $y - k = \pm \frac{b}{a}(x - h)$ . The asymptotes guide the shape of the branches as they extend toward infinity.
What do $a$ and $b$ represent, and how do they differ from an ellipse?: In this hyperbola form, $a$ is the semi-transverse axis length—the distance from the center to each vertex along the horizontal direction. The value $b$ is the semi-conjugate axis length, which does not correspond to any point on the hyperbola itself, but instead controls the steepness of the asymptotes via the slope $\frac{b}{a}$ . In an ellipse, both $a$ and $b$ represent points on the curve; in a hyperbola, only $a$ does.
Can I shift or rotate a hyperbola like other curves?: Yes, the $(h, k)$ in this equation shifts the center away from the origin, and this form represents a hyperbola with axes parallel to the coordinate axes. Rotated hyperbolas exist but require a more complex equation that includes an $xy$ term, which is beyond the scope of this standard form.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18