Eccentricity of Hyperbola - Definition, Formula, and Examples

What Is the Eccentricity of a Hyperbola?

The eccentricity of a hyperbola is the ratio of the distance from the centre to a focus, divided by the distance from the centre to a vertex. It is written $e = \dfrac{c}{a}$, where $c$ is the focal distance and $a$ is the length of the semi-transverse axis. Because a hyperbola always has its foci farther from the centre than its vertices, $c > a$, so the ratio is always more than 1.

Eccentricity belongs to a single idea that runs through every conic: the focus-directrix ratio $\dfrac{PF}{PM} = e$. When $e < 1$ the curve closes into an ellipse, when $e = 1$ it opens into a parabola, and when $e > 1$ it splits into the two branches of a hyperbola. So eccentricity is the one dial that turns one conic into another.

The Number That Told Astronomers a Visitor Came From Another Star

In 2017, a strange object swept past the Sun and left again, never to return. Astronomers tracked its path and found its orbit was not a closed loop but an open curve, a hyperbola, with an eccentricity well above 1. That single number proved ʻOumuamua was not bound to our Sun at all: it had drifted in from interstellar space. The eccentricity of a hyperbola is exactly this measure, a number that says how sharply an open curve flares away from its centre.

What Is the Formula For the Eccentricity of a Hyperbola?

For the standard hyperbola with a horizontal transverse axis,

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

the eccentricity is

$$e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} = \sqrt{1 + \frac{b^2}{a^2}}, \qquad e > 1.$$

The variables mean: $a$ is the semi-transverse axis (centre to vertex), $b$ is the semi-conjugate axis, and $c$ is the focal distance (centre to focus), tied together by the hyperbola relation $c^2 = a^2 + b^2$. The same formula holds for a vertical hyperbola $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$, with $a$ still the semi-transverse axis under the positive term.

Note the sign carefully: the hyperbola uses $c^2 = a^2 + b^2$ (a plus), while the foci of a hyperbola sit outside the vertices because of it. The ellipse uses $c^2 = a^2 - b^2$ (a minus). That single sign is what pushes hyperbola eccentricity above 1 and keeps ellipse eccentricity below it.

Where Does the Formula e = c/a Come From?

The eccentricity is not an extra definition bolted on; it drops straight out of the focus-directrix rule applied at a vertex. Take the right vertex $(a, 0)$, which lies on the curve, so it obeys $\dfrac{PF}{PM} = e$.

Its distance to the right focus $F(c, 0)$ is

$$PF = c - a.$$

Its perpendicular distance to the right directrix $x = \dfrac{a}{e}$ is

$$PM = a - \frac{a}{e}.$$

Set the ratio equal to $e$:

$$\frac{c - a}{a - \frac{a}{e}} = e$$

Multiply out the denominator:

$$c - a = e\left(a - \frac{a}{e}\right) = ea - a$$

$$c = ea$$

$$e = \frac{c}{a}$$

Since $c^2 = a^2 + b^2$ gives $c = \sqrt{a^2 + b^2} > a$, the ratio $\dfrac{c}{a}$ is always greater than 1. That is the algebraic reason a hyperbola's eccentricity can never dip to 1 or below.

Why Is the Eccentricity of a Hyperbola Always Greater Than 1?

Because the foci always lie beyond the vertices. In a hyperbola, $c^2 = a^2 + b^2$, and since $b^2 > 0$, we get $c^2 > a^2$, so $c > a$ and therefore $\dfrac{c}{a} > 1$. There is no way to arrange a real hyperbola with the focus inside the vertex.

Contrast this with the closed conics. A circle has coincident foci, so $c = 0$ and $e = 0$. An ellipse keeps its foci inside the curve, so $c < a$ and $0 < e < 1$. Only when the defining relation flips to a plus sign do the foci escape past the vertices, and the eccentricity climbs past 1. The larger $\dfrac{b}{a}$ is, the wider the branches open and the larger $e$ grows.

Examples of Eccentricity of Hyperbola

Six examples, from reading $e$ off a standard equation to recovering a hyperbola from its eccentricity.

Example 1

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

Here $a^2 = 9$ and $b^2 = 16$, so $a = 3$. Find $c$, then $e$:

$c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$e = \dfrac{c}{a} = \dfrac{5}{3}$

Final answer: $e = \dfrac{5}{3} \approx 1.67$. The first instinct on these problems is to reach for $a$ under the term that looks bigger; always take $a^2$ as the denominator under the positive term.

Example 2

A student finds the eccentricity of $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ by writing "$e = \sqrt{1 - \tfrac{b^2}{a^2}} = \sqrt{1 - \tfrac{16}{9}}$." Where does this go wrong?

The tempting move borrows the ellipse formula, which subtracts $\dfrac{b^2}{a^2}$.

Under the root that gives $1 - \dfrac{16}{9} = -\dfrac{7}{9}$, a negative number. The square root of a negative is not a real eccentricity, and that impossible result is the signal the wrong formula was used.

The correct method uses the hyperbola relation $c^2 = a^2 + b^2$, which adds:

$e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{16}{9}} = \sqrt{\dfrac{25}{9}} = \dfrac{5}{3}$

Final answer: $e = \dfrac{5}{3}$. Ellipse subtracts, hyperbola adds; the negative under the root is the tell.

Example 3

Find the eccentricity of $\dfrac{x^2}{49} - \dfrac{y^2}{25} = 1$.

Here $a^2 = 49$, $b^2 = 25$, so $a = 7$.

$c = \sqrt{49 + 25} = \sqrt{74}$

$e = \dfrac{\sqrt{74}}{7} \approx 1.23$

Final answer: $e = \dfrac{\sqrt{74}}{7}$. Not every eccentricity is a tidy fraction; leaving it as an exact radical is the precise form.

Example 4

A hyperbola has $a = 4$ and eccentricity $e = \dfrac{5}{4}$. Find its equation.

From $e = \dfrac{c}{a}$, first recover $c$:

$c = ea = \dfrac{5}{4} \times 4 = 5$

Then use $c^2 = a^2 + b^2$ to find $b^2$:

$b^2 = c^2 - a^2 = 25 - 16 = 9$

Final answer: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$. Working backward, eccentricity fixes $c$, and the plus-sign relation fixes $b$.

Example 5

Find the eccentricity of the rectangular hyperbola $x^2 - y^2 = 16$.

Write it in standard form by dividing through by 16:

$\dfrac{x^2}{16} - \dfrac{y^2}{16} = 1$

Here $a^2 = b^2 = 16$, so $a = b = 4$.

$c = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

$e = \dfrac{4\sqrt{2}}{4} = \sqrt{2}$

Final answer: $e = \sqrt{2} \approx 1.41$. Every rectangular (equilateral) hyperbola, where $a = b$, has this same eccentricity.

Example 6

A hyperbola has foci at $(\pm 13, 0)$ and vertices at $(\pm 5, 0)$. Find its eccentricity and the equation.

Read the pieces directly: $c = 13$ (focus) and $a = 5$ (vertex).

$e = \dfrac{c}{a} = \dfrac{13}{5} = 2.6$

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

Final answer: $e = \dfrac{13}{5}$ and $\dfrac{x^2}{25} - \dfrac{y^2}{144} = 1$. The reasoning step is spotting that foci give $c$ and vertices give $a$, not the other way round.

Why the Eccentricity of a Hyperbola Matters: It Reads the Shape of an Orbit

Eccentricity is not a decorative label; it is the single number that decides whether a path returns or escapes. For a curve, it is the geometric shorthand for "how open."

• It classifies the conic. With $e > 1$ you have a hyperbola, with $e = 1$ a parabola, and with $e < 1$ an ellipse. One number, three curves.

• It sets how wide the branches flare. Because $e = \sqrt{1 + \tfrac{b^2}{a^2}}$, a larger conjugate axis $b$ opens the branches wider and raises $e$; an eccentricity just above 1 gives a nearly parabola-like curve.

• It fixes the asymptote slope. The asymptotes $y = \pm\tfrac{b}{a}x$ are tied to the same $\tfrac{b}{a}$ that drives eccentricity, so $e$ and the asymptotes carry the same information.

What Are the Most Common Mistakes With the Eccentricity of a Hyperbola?

Three errors account for most wrong answers, and two of them come from confusing the hyperbola with the ellipse.

Mistake 1: Using the ellipse sign in the c relation

Where it slips in: Recalling one "$c^2 = a^2 \pm b^2$" relation and guessing the sign.

Don't do this: Writing $c^2 = a^2 - b^2$ for a hyperbola.

The correct way: The hyperbola uses $c^2 = a^2 + b^2$ (add), the ellipse uses $c^2 = a^2 - b^2$ (subtract). Students first meeting conics carry the ellipse habit across because both curves use $a$, $b$, and $c$. The check that settles it: a hyperbola's foci sit outside its vertices, so $c$ must be bigger than $a$, which only the plus sign gives.

Mistake 2: Taking a² from the wrong term

Where it slips in: Assuming the larger denominator is always $a^2$.

Don't do this: For $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$, treating $16$ as $a^2$ because it is larger.

The correct way: In a hyperbola, $a^2$ is the denominator under the positive term, regardless of which number is larger. Here the positive term is $\dfrac{x^2}{9}$, so $a^2 = 9$. The rusher grabs the bigger number; the transverse axis is set by the sign, not the size.

Mistake 3: Expecting e between 0 and 1

Where it slips in: Porting the ellipse's range to the hyperbola.

Don't do this: Reporting an eccentricity like $0.8$ for a hyperbola and moving on.

The correct way: A hyperbola always has $e > 1$. Any answer of 1 or below means an arithmetic slip or the wrong conic. The confusion between the closed-curve range ($e < 1$) and the open-curve range ($e > 1$) is the single most common source of wrong answers here.

Conclusion

• The eccentricity of a hyperbola is $e = \dfrac{c}{a} = \sqrt{1 + \dfrac{b^2}{a^2}}$, and it is always greater than 1.

• The defining relation is $c^2 = a^2 + b^2$ (add), which is what pushes the foci outside the vertices.

• Take $a^2$ from the denominator under the positive term, never simply the larger number.

• A larger $\dfrac{b}{a}$ opens the branches wider and raises the eccentricity; a rectangular hyperbola always has $e = \sqrt{2}$.

To take conic sections further with a teacher, explore Bhanzu's geometry tutor, a high school math tutor, or math classes online.

Practice These to Solidify Your Understanding

Work through these, then check your answers:

1. Find the eccentricity of $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$. (Answer to Question 1: $a = 4$, $c = 5$, so $e = \tfrac{5}{4}$.)

2. A hyperbola has $a = 6$ and $e = \tfrac{5}{3}$. Find $b^2$. (Answer to Question 2: $c = 10$, so $b^2 = 100 - 36 = 64$.)

3. Find the eccentricity of $\dfrac{y^2}{4} - \dfrac{x^2}{5} = 1$. (Answer to Question 3: $a^2 = 4$, $c^2 = 9$, so $e = \tfrac{3}{2}$.)

If Question 3 tripped you, revisit the note that $a^2$ sits under the positive term. Want a trainer to walk conics through with your child? Book a free demo class.

Read More

• Eccentricity of an ellipse — the same ratio for a closed curve, where $e < 1$.

• Vertex of a hyperbola — the turning points at $(\pm a, 0)$ that set the value of $a$.

• Eccentricity — the single number that classifies every conic section.

• Directrix of an ellipse — the line paired with a focus through the ratio $e$.

• Parabola — the borderline conic where $e = 1$.