In short

A hyperbola is the set of all points whose distances from two fixed points (the foci) have a constant difference. Its standard equation is \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1, and it consists of two separate branches. The number e = c/a (where c is the distance from the centre to each focus) is the eccentricity, and for a hyperbola e > 1. The transverse axis connects the two vertices, the conjugate axis is perpendicular to it, and the relation connecting the parameters is c^2 = a^2 + b^2 — a plus, not the minus you had for the ellipse.

Think of a sonic boom. A fighter jet flying faster than sound creates a shock wave that expands outward as a cone. If you slice that cone with a horizontal plane, the cross-section you get is not a circle, not an ellipse, not a parabola — it is a hyperbola. Two separate curves, opening away from each other, spreading outward without limit.

You have already met the ellipse, where a point moves so that the sum of its distances from two foci is constant. The hyperbola is the mirror version: a point moves so that the difference of its distances from two foci is constant. The sum locks the point inside a closed oval. The difference lets the point escape to infinity — in two opposite directions. That single sign change, from + to -, transforms the geometry completely.

The hyperbola is the last of the three non-degenerate conic sections you will study (after the parabola and the ellipse). It shows up in the paths of comets that visit the solar system once and never return (they follow hyperbolic orbits around the Sun), in the shape of cooling towers at thermal power plants, and in the mathematics of navigation systems that locate a ship by measuring the time difference of signals from two stations. The two-station navigation problem is literally the definition of a hyperbola.

The definition, precisely

Definition of a hyperbola

A hyperbola is the locus of a point P in the plane such that the absolute difference of its distances from two fixed points F_1 and F_2 (the foci) is a constant:

|PF_1 - PF_2| = 2a

where 2a is a positive constant less than the distance F_1F_2 = 2c between the foci.

The notation parallels the ellipse:

The absolute value |PF_1 - PF_2| is essential. Without it, the condition PF_1 - PF_2 = 2a (no absolute value) picks out only one of the two branches — the branch closer to F_2. With the absolute value, you get both branches.

Compare with the ellipse: for the ellipse, PF_1 + PF_2 = 2a with a > c. For the hyperbola, |PF_1 - PF_2| = 2a with a < c. Sum versus difference, a > c versus a < c. Everything else follows from this sign change.

The defining property: for any point $P$ on the hyperbola, $|PF_1 - PF_2| = 2a = 6$. Here $a = 3$, $b = 4$, $c = 5$. The point $P \approx (4, 4.22)$ is on the right branch, so $PF_1 - PF_2 = 2a = 6$. On the left branch, the inequality reverses: $PF_2 - PF_1 = 6$.

Deriving the standard equation

Place the centre at the origin. Put the foci on the x-axis at F_1(-c, 0) and F_2(c, 0). Take any point P(x, y) on the right branch of the hyperbola (so P is closer to F_2). The defining condition, for this branch, is:

PF_1 - PF_2 = 2a

Write out the distances:

\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a

The derivation follows the same strategy as the ellipse — isolate one radical and square, twice.

Step 1. Isolate one radical.

\sqrt{(x + c)^2 + y^2} = 2a + \sqrt{(x - c)^2 + y^2}

Why: move the second radical to the right so you can square both sides. Notice the +2a on the right — for the ellipse it was 2a minus the second radical, here it is plus. This is the sign difference that changes everything.

Step 2. Square both sides.

(x + c)^2 + y^2 = 4a^2 + 4a\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2

Expand (x + c)^2 = x^2 + 2cx + c^2 and (x - c)^2 = x^2 - 2cx + c^2. The x^2, y^2, and c^2 terms cancel:

2cx = 4a^2 + 4a\sqrt{(x - c)^2 + y^2} - 2cx
4cx - 4a^2 = 4a\sqrt{(x - c)^2 + y^2}

Divide by 4:

cx - a^2 = a\sqrt{(x - c)^2 + y^2}

Why: after expanding and canceling, you are left with one radical isolated on one side. For the ellipse the same step gave a^2 - cx = a\sqrt{(x-c)^2 + y^2}. The sign is flipped because the defining equation had a minus instead of a plus.

Step 3. Square again.

(cx - a^2)^2 = a^2[(x - c)^2 + y^2]
c^2 x^2 - 2a^2 cx + a^4 = a^2 x^2 - 2a^2 cx + a^2 c^2 + a^2 y^2

The -2a^2 cx terms cancel:

c^2 x^2 + a^4 = a^2 x^2 + a^2 c^2 + a^2 y^2
c^2 x^2 - a^2 x^2 - a^2 y^2 = a^2 c^2 - a^4
x^2(c^2 - a^2) - a^2 y^2 = a^2(c^2 - a^2)

Why: collect x^2 terms on the left and constants on the right. The expression c^2 - a^2 appears on both sides — and since c > a, this quantity is positive.

Step 4. Define b^2 = c^2 - a^2.

Since c > a, we have b > 0. Substitute:

x^2 b^2 - a^2 y^2 = a^2 b^2

Divide by a^2 b^2:

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

Why: dividing by a^2 b^2 normalises the equation to the cleanest possible form. The constant on the right is 1, and the minus sign between the two terms is the hallmark of the hyperbola.

Standard form of a hyperbola

The equation of a hyperbola with centre at the origin and foci on the x-axis is

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

where a > 0, b > 0, the foci are at (\pm c, 0) with c = \sqrt{a^2 + b^2}, and the absolute difference of focal distances is 2a.

If the foci are on the y-axis instead, the equation becomes \dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1.

Compare this with the ellipse equation \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1. The only algebraic change is + to -. But geometrically, the curve splits open: the ellipse is a single closed loop, the hyperbola is two separate branches extending to infinity.

The key relation is c^2 = a^2 + b^2 — a plus sign, not the c^2 = a^2 - b^2 of the ellipse. For the ellipse, c < a and the foci are inside the curve. For the hyperbola, c > a and the foci are outside the vertices.

The anatomy of a hyperbola

Every term has a geometric meaning. Here they are, labelled on one picture.

The hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$, with $a = 3$, $b = 4$, $c = 5$. The foci (red) are at $(\pm 5, 0)$, the vertices at $(\pm 3, 0)$. The transverse axis connects the vertices; the conjugate axis is perpendicular to it. Note that $B(0, 4)$ and $B'(0, -4)$ are *not* on the hyperbola — they mark the ends of the conjugate axis.

Unlike the ellipse, a and b are not required to satisfy a > b. For a hyperbola, a and b can be in any ratio. The transverse axis is simply the axis along which the curve opens — the one containing the foci and vertices.

The relation c^2 = a^2 + b^2 means: if you draw a right triangle with legs a and b, the hypotenuse is c. For the ellipse, a was the hypotenuse and b, c were the legs. For the hyperbola, c is the hypotenuse. The Pythagorean triple (3, 4, 5) appears in the example above: a = 3, b = 4, c = 5.

Eccentricity

The eccentricity of a hyperbola is the ratio

e = \frac{c}{a}

For a hyperbola, c > a, so e > 1. This is the defining feature that separates it from the ellipse (0 < e < 1) and the parabola (e = 1).

Since c^2 = a^2 + b^2, you can write e = \sqrt{1 + b^2/a^2}, or equivalently, b^2 = a^2(e^2 - 1).

Four hyperbolas with $a = 3$ but different eccentricities. Dashed: $e \approx 1.03$ ($b \approx 0.75$, nearly flat branches). Thin solid: $e \approx 1.33$ ($b \approx 2.65$). Bold solid: $e = 5/3 \approx 1.67$ ($b = 4$). Red solid: $e = 3$ ($b \approx 8.49$, narrow branches). As $e$ increases, the branches tighten.

Eccentricity and the conic family

The eccentricity e is the single number that classifies every conic section:

Eccentricity Conic
e = 0 Circle
0 < e < 1 Ellipse
e = 1 Parabola
e > 1 Hyperbola

As e increases through 1, the closed ellipse opens into a parabola (which opens on one side) and then into a hyperbola (which opens on two sides). The three curves form a continuous family, controlled by a single parameter.

The focus-directrix definition

Like the ellipse and parabola, the hyperbola has a focus-directrix definition.

Focus-directrix definition

A hyperbola is the locus of a point P such that

\frac{\text{distance from } P \text{ to a focus}}{\text{distance from } P \text{ to the corresponding directrix}} = e > 1

For the standard hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, each focus has a corresponding directrix:

Since e > 1, the directrix x = a/e lies inside the hyperbola (between the two vertices: a/e < a, so the directrix is closer to the centre than the vertex). This is the opposite of the ellipse, where the directrix lies outside.

Verification. Take a point (x, y) on the right branch. Its distance from F_2(c, 0) is \sqrt{(x-c)^2 + y^2}. From the derivation, cx - a^2 = a\sqrt{(x-c)^2 + y^2}, so the distance is (cx - a^2)/a = ex - a. The distance from the directrix x = a/e is x - a/e. The ratio is:

\frac{ex - a}{x - a/e} = \frac{ex - a}{(ex - a)/e} = e

This confirms the equivalence.

The hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ with its two directrices at $x = \pm a^2/c = \pm 9/5 = \pm 1.8$ (dotted lines). The directrices lie between the two vertices, inside the hyperbola. For every point $P$ on the curve, the ratio $PF/Pd$ (distance to focus divided by distance to directrix) equals $e = 5/3$.

The latus rectum

The latus rectum is the chord through a focus perpendicular to the transverse axis. Its computation follows the same pattern as the ellipse.

Take the focus F_2(c, 0). The latus rectum has x = c. Substitute into the hyperbola:

\frac{c^2}{a^2} - \frac{y^2}{b^2} = 1
\frac{y^2}{b^2} = \frac{c^2}{a^2} - 1 = \frac{c^2 - a^2}{a^2} = \frac{b^2}{a^2}
y = \pm\frac{b^2}{a}

Latus rectum of a hyperbola

The length of the latus rectum of the hyperbola \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 is

\boxed{\ell = \frac{2b^2}{a}}

The formula is identical to the ellipse. The derivation is identical too — the difference in signs in the equation does not affect this particular computation, because c^2 - a^2 = b^2 plays the same role as a^2 - c^2 = b^2 in the ellipse derivation.

Worked examples

Example 1: Finding the equation from the foci and eccentricity

A hyperbola has foci at (\pm 13, 0) and eccentricity e = 13/5. Find its equation, vertices, and the length of the latus rectum.

Step 1. Identify c and e. The foci are at (\pm 13, 0), so c = 13. The eccentricity is e = 13/5.

Why: since the foci are on the x-axis and symmetric about the origin, the standard form \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 applies.

Step 2. Find a.

e = \frac{c}{a} \implies a = \frac{c}{e} = \frac{13}{13/5} = 5

Step 3. Find b.

b^2 = c^2 - a^2 = 169 - 25 = 144, \quad b = 12

Why: for the hyperbola, the relation is c^2 = a^2 + b^2, so b^2 = c^2 - a^2. This is the opposite of the ellipse relation b^2 = a^2 - c^2.

Step 4. Write the equation.

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

Step 5. Compute the latus rectum.

\ell = \frac{2b^2}{a} = \frac{2 \times 144}{5} = \frac{288}{5} = 57.6

Result: The hyperbola is \frac{x^2}{25} - \frac{y^2}{144} = 1, with vertices at (\pm 5, 0) and latus rectum of length 288/5.

The hyperbola $\frac{x^2}{25} - \frac{y^2}{144} = 1$, with $a = 5$, $b = 12$, $c = 13$ (a 5-12-13 Pythagorean triple). The foci are far from the vertices, giving a large eccentricity of $13/5 = 2.6$. The latus rectum through $F_2$ is drawn as a dashed vertical line.

The numbers (5, 12, 13) form a Pythagorean triple, so c = \sqrt{a^2 + b^2} = \sqrt{25 + 144} = \sqrt{169} = 13 comes out clean. In exam problems, look for Pythagorean triples — they appear frequently in conic-section questions.

Example 2: Identifying the conic from its equation

Identify the conic 9x^2 - 16y^2 = 144. Find its centre, foci, vertices, eccentricity, directrices, and latus rectum.

Step 1. Divide by 144 to put the equation in standard form.

\frac{x^2}{16} - \frac{y^2}{9} = 1

This is a hyperbola (minus sign between the terms) with a^2 = 16 and b^2 = 9.

Why: dividing by 144 reduces both coefficients and reveals the standard form. The term with the positive sign has a^2 under it.

Step 2. Extract a, b, c.

a = 4, \quad b = 3, \quad c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Another Pythagorean triple: (3, 4, 5) with a = 4, b = 3, c = 5.

Step 3. Find the eccentricity and directrices.

e = \frac{c}{a} = \frac{5}{4} = 1.25
\text{Directrices: } x = \pm\frac{a}{e} = \pm\frac{4}{5/4} = \pm\frac{16}{5} = \pm 3.2

Why: the directrices are at x = \pm a^2/c = \pm a/e. Since e > 1, the value a/e < a, so the directrices sit between the vertices — inside the hyperbola.

Step 4. Compute the latus rectum.

\ell = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2} = 4.5

Result: Centre (0, 0); vertices (\pm 4, 0); foci (\pm 5, 0); e = 5/4; directrices x = \pm 16/5; latus rectum = 9/2.

The hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ with $a = 4$, $b = 3$, $c = 5$. The directrices (dotted) are at $x = \pm 3.2$, sitting between the two vertices. The latus rectum through $F_2$ has endpoints at $(5, \pm 9/4)$.

Every quantity follows from the standard form. Once you identify a^2, b^2, and compute c, all the other features — eccentricity, foci, directrices, latus rectum — are one formula away.

Common confusions

Going deeper

If you came here for the definition, standard form, and key terms of the hyperbola, you have them. The rest of this article covers the rectangular hyperbola, the connection to the other conics via the directrix, and the two-focus construction.

The rectangular hyperbola

When a = b, the hyperbola \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 simplifies to x^2 - y^2 = a^2. Its eccentricity is e = \sqrt{1 + 1} = \sqrt{2}. This is called a rectangular hyperbola because its asymptotes (the lines y = \pm x) are perpendicular to each other.

If you rotate a rectangular hyperbola by 45°, its equation becomes xy = k for some constant k. The familiar curve y = 1/x — the hyperbola from the derivative article — is a rectangular hyperbola with k = 1. So the function you have been differentiating since you first met derivatives is a conic section in disguise.

The conic family, unified

All three non-degenerate conics — ellipse, parabola, hyperbola — can be described by a single equation using the focus-directrix definition:

\frac{\text{distance to focus}}{\text{distance to directrix}} = e

The parabola is the boundary case. As e approaches 1 from below, the ellipse stretches longer and longer until one end escapes to infinity — becoming a parabola. As e increases past 1, the curve "breaks" into two pieces — the parabola's single branch splits into the hyperbola's two branches. The entire family is controlled by this single number e.

Bhaskara II, in the 12th-century Siddhanta Shiromani, worked with the geometry of planetary orbits that we now recognise as conic sections. The unifying role of eccentricity, linking circles to ellipses to parabolas to hyperbolas, was not formalised until later — but the geometric intuition that these curves form a family was present in Indian astronomical traditions long before the modern algebraic framework.

The two-focus construction

For the ellipse, you used the "string and pins" construction: two pins (foci), a loop of string, and a pencil tracing all points where the sum of distances is constant. For the hyperbola, the analogous construction uses a difference of distances — harder to do with string, but possible with a mechanical device.

Take two pins at F_1 and F_2. Attach a ruler of length \ell to F_1 at one end, so it can rotate freely. Tie a string of length \ell - 2a from the other end of the ruler to F_2. Now hold a pencil where the string meets the ruler, keeping the string taut. As the ruler rotates, the pencil traces a curve. At each position, the pencil is at distance r_1 from F_1 (along the ruler) and r_2 from F_2 (along the string). The ruler has length \ell = r_1 + (\ell - 2a - r_2 + r_2)... more simply: the string length is \ell - 2a, so r_2 = \ell - r_1 - (\ell - 2a - r_2). The constraint is r_1 - r_2 = 2a, which traces one branch of the hyperbola.

This is a physical construction, not just algebra. If you have a ruler, string, and two pins, you can draw a hyperbola on a board — the same way you drew an ellipse, but with the defining condition changed from sum to difference.

Where this leads next

You now know what a hyperbola is, where its equation comes from, and what each term means. The next articles develop the tools for working with it: