Hyperbola: Definition, Equation, Foci, and Asymptotes

#Geometry
TL;DR
A hyperbola is the set of all points whose distances to two fixed points (the foci) differ by a constant — giving two open curves that mirror each other. This guide covers the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the foci, vertices, asymptotes, and eccentricity, with the relation $c^2 = a^2 + b^2$, worked examples, and the slips to watch for.
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Bhanzu TeamLast updated on July 13, 20269 min read

What Is a Hyperbola?

A hyperbola is the set of all points in a plane for which the difference of the distances to two fixed points, called the foci, is a positive constant. Those two distances need not be equal — only their difference stays fixed, and that constant difference is what bends the curve into two separate branches.

A hyperbola is one of the four conic sections — the curves you get by slicing a cone with a plane. Slice gently and you get a circle or an ellipse; slice parallel to the side and you get a parabola; slice steeply enough to cut both halves of the cone and you get a hyperbola's two branches.

The contrast with the ellipse is the cleanest way to hold the idea. An ellipse is the set of points where the sum of the two focal distances is constant — a closed loop. Swap "sum" for "difference" and the loop breaks open into two mirror curves. That single word, difference versus sum, is the whole distinction.

How Is A Hyperbola Different From A Parabola?

Both curves open outward, so they are easy to confuse. A parabola is a single open curve, defined by one focus and a directrix line; a hyperbola has two branches and two foci. A parabola has no asymptotes; a hyperbola's branches hug two straight asymptote lines as they run off to infinity.

The Standard Equation of a Hyperbola

For a hyperbola centred at the origin with its transverse axis (the axis through both vertices) along the x-axis, the standard equation is:

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

If the transverse axis runs along the y-axis instead, the equation flips:

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

The single feature that tells the two apart is which variable carries the positive term. Positive $x^2$ term means the branches open left and right; positive $y^2$ term means they open up and down.

Variable key: $a$ is the distance from the centre to each vertex (so the vertices sit at $(\pm a, 0)$ for a horizontal hyperbola); $b$ sets the height of the central rectangle that controls the asymptotes; $c$ is the distance from the centre to each focus.

The focus relation, derived

For a hyperbola, the three lengths are tied together by:

$$c^2 = a^2 + b^2$$

This looks like the ellipse relation but with a plus sign instead of a minus — and the plus is not arbitrary. It falls straight out of the definition. Take the vertex point $(a, 0)$, which lies on the curve. Its distance to the near focus is $c - a$ and to the far focus is $c + a$, so the constant difference of focal distances is $(c + a) - (c - a) = 2a$. Working that constant through the distance definition for a general point and simplifying produces $c^2 = a^2 + b^2$ — which is why, for a hyperbola, the focus always sits farther from the centre than the vertex.

What does the value of $b$ control if the vertices only depend on $a$? Good question — $b$ never touches the curve directly, since the branches cross the axis only at the vertices $(\pm a, 0)$. Instead $b$ sets the height of the central rectangle, and that rectangle fixes the slope of the asymptotes. So $a$ controls where the curve starts and $b$ controls how steeply it spreads.

The Key Parts of a Hyperbola

Part

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Centre

$(0, 0)$

Vertices

$(\pm a, 0)$

Foci

$(\pm c, 0)$, where $c = \sqrt{a^2 + b^2}$

Transverse axis length

$2a$

Conjugate axis length

$2b$

Asymptotes

$y = \pm \dfrac{b}{a} x$

Eccentricity

$e = \dfrac{c}{a} = \sqrt{1 + \dfrac{b^2}{a^2}}$

The asymptotes are the two straight lines the branches approach but never touch. The fastest way to draw them: mark the central rectangle of width $2a$ and height $2b$, then extend its diagonals. The eccentricity $e$ measures how "open" the curve is; for every hyperbola $e > 1$, and the larger $e$, the wider the branches spread.

Examples of Hyperbola

The examples move from reading values off a standard equation to building one from given parts. Coordinates are in plain units throughout.

Example 1

For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the vertices.

Here $a^2 = 9$, so $a = 3$. The positive term is $x^2$, so the hyperbola opens horizontally and the vertices sit on the x-axis:

$$\text{Vertices: } (\pm 3, 0)$$

Example 2

For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the foci — and avoid the most common slip.

A reader who has just met the ellipse often reaches for $c^2 = a^2 - b^2$ out of habit, giving $c^2 = 9 - 16 = -7$. A squared length cannot be negative, so the result is impossible — the formula was the wrong one. That impossibility is the signal to stop.

For a hyperbola the relation carries a plus sign:

$$c^2 = a^2 + b^2 = 9 + 16 = 25, \quad c = 5$$

$$\text{Foci: } (\pm 5, 0)$$

Example 3

For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the asymptotes.

With $a = 3$ and $b = 4$, the asymptotes are:

$$y = \pm \frac{b}{a} x = \pm \frac{4}{3} x$$

Example 4

Find the eccentricity of $\frac{x^2}{9} - \frac{y^2}{16} = 1$.

Using $c = 5$ from Example 2 and $a = 3$:

$$e = \frac{c}{a} = \frac{5}{3} \approx 1.67$$

Since $e > 1$, this confirms the curve is a hyperbola.

Example 5

A hyperbola opens up and down with vertices at $(0, \pm 5)$ and foci at $(0, \pm 13)$. Find its equation.

The vertices are on the y-axis, so the $y^2$ term is positive and $a = 5$. The foci give $c = 13$. Solve for $b^2$:

$$b^2 = c^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144$$

$$\frac{y^2}{25} - \frac{x^2}{144} = 1$$

Example 6

The difference of the distances from a point to two stations at $(\pm 10, 0)$ km is always 12 km. Find the hyperbola the point lies on.

The constant difference equals $2a$, so $2a = 12$ gives $a = 6$ and $a^2 = 36$. The stations are the foci, so $c = 10$ and:

$$b^2 = c^2 - a^2 = 100 - 36 = 64$$

$$\frac{x^2}{36} - \frac{y^2}{64} = 1$$

This is exactly the navigation problem from the start of the article, written as an equation.

Why the Hyperbola Matters

The hyperbola is the geometry of differences, and differences turn out to run the physical world.

  • Navigation and positioning — systems that locate a receiver from the difference in signal arrival times place it on a hyperbola; this is the principle behind hyperbolic navigation.

  • Astronomy — an object passing the Sun with more than escape speed follows a hyperbolic orbit, swinging by once and never returning. Comets on hyperbolic paths are visitors, not residents.

  • Optics and design — hyperbolic mirrors redirect light toward a focus, a property used in some reflecting telescopes.

The reason the curve keeps appearing is structural: whenever a quantity is governed by a constant difference rather than a constant sum, the geometry is a hyperbola. The conic sections were studied in depth by the Greek geometer Apollonius of Perga around 200 BCE — he gave the hyperbola, ellipse, and parabola their names, centuries before anyone knew planets and comets would trace them.

Where Hyperbola Problems Go Wrong

Mistake 1: Using the ellipse focus relation

Where it slips in: Right after studying the ellipse, when $c^2 = a^2 - b^2$ is fresh in memory.

Don't do this: Apply $c^2 = a^2 - b^2$ to a hyperbola — it can produce a negative number for $c^2$, which is impossible.

The correct way: A hyperbola uses $c^2 = a^2 + b^2$ (plus, not minus), because the focus sits farther from the centre than the vertex. The first-instinct error is carrying the ellipse's minus sign over — and the negative $c^2$ it produces is the tell-tale sign you reached for the wrong relation.

Mistake 2: Misreading which way the hyperbola opens

Where it slips in: Identifying vertices and foci from the equation.

Don't do this: Assume the curve always opens left-right, or read $a$ from whichever denominator is larger.

The correct way: The variable with the positive term sets the opening direction and carries $a^2$. In $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the positive $y^2$ term means it opens vertically, even if $b^2 > a^2$. For a hyperbola, unlike an ellipse, $a$ is not "the bigger one" — it is "the one under the positive term".

Mistake 3: Flipping the asymptote slope

Where it slips in: Writing the asymptotes for a horizontal hyperbola.

Don't do this: Write $y = \pm \frac{a}{b} x$ from memory.

The correct way: For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the slopes are $\pm \frac{b}{a}$ — rise over run from the central rectangle, where the rectangle is $2a$ wide and $2b$ tall. The second-guesser who drew the rectangle correctly often distrusts it and inverts the fraction at the last step; the rectangle is the answer, so trust it.

Conclusion

  • A hyperbola is the set of points whose distances to two foci have a constant difference, producing two mirror-image branches.

  • Its standard equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, with the positive term setting the opening direction.

  • The focus relation is $c^2 = a^2 + b^2$ — a plus sign, unlike the ellipse.

  • Asymptotes are $y = \pm \frac{b}{a}x$, and eccentricity $e = \frac{c}{a} > 1$ for every hyperbola.

  • The most common mistake is borrowing the ellipse's minus-sign focus relation, which gives an impossible negative $c^2$.

Practice and Next Steps

Work through these problems to solidify your understanding, then verify each against the formulas above.

  1. Find the vertices, foci, and asymptotes of $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

  2. A hyperbola has vertices $(\pm 4, 0)$ and eccentricity $\frac{5}{4}$. Find its equation.

  3. Find the eccentricity of $\frac{y^2}{36} - \frac{x^2}{64} = 1$.

To work through conic sections with a teacher who derives each relation rather than handing you a formula sheet, explore Bhanzu's geometry tutor, our high school math tutor, or math tutoring. Want a live Bhanzu trainer to walk through more hyperbola problems? Book a free demo class.

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Frequently Asked Questions

Is a hyperbola two separate curves or one?
It is one hyperbola made of two branches. The two branches together form a single curve defined by one equation and one pair of foci.
What is the eccentricity of a hyperbola?
Eccentricity is $e = \frac{c}{a}$, and for every hyperbola it is greater than 1. A circle has $e = 0$, an ellipse has $0 < e < 1$, a parabola has $e = 1$, and a hyperbola has $e > 1$.
Does a hyperbola ever touch its asymptotes?
No. The branches get arbitrarily close to the asymptotes as they run outward but never meet them — the asymptotes are guide lines, not part of the curve.
What is the rectangular (equilateral) hyperbola?
When $a = b$, the asymptotes are perpendicular and the curve is called a rectangular hyperbola. The familiar graph of $xy = k$ is a rectangular hyperbola rotated onto the axes.
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