Foci of Hyperbola — Formula, How to Find, and Examples

What Are The Foci Of A Hyperbola?

The foci of a hyperbola are two fixed points, $F_1$ and $F_2$, that define the curve through a difference of distances. For every point $P$ on the hyperbola, the absolute difference between its distances to the two foci is a constant:

$$|PF_1 - PF_2| = 2a$$

That constant equals $2a$, twice the distance from the center to a vertex. This difference-of-distances rule is the defining feature of a hyperbola — and it is the mirror image of the ellipse, where the sum of distances to the two foci of the ellipse is constant. Sum gives an ellipse; difference gives a hyperbola.

The foci always lie inside the two branches, further from the center than the vertices, along the axis the hyperbola opens along.

How Do You Find The Foci Of A Hyperbola?

For a hyperbola in standard form centered at the origin, the foci are found from one relationship. Take the horizontal hyperbola:

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

The distance $c$ from the center to each focus satisfies:

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

So $c = \sqrt{a^2 + b^2}$, and the foci sit at $(\pm c, 0)$. For a vertical hyperbola, $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$, the same $c$ applies and the foci are at $(0, \pm c)$.

Notice the plus sign. For a hyperbola, $c$ is larger than $a$ because $c^2 = a^2 + b^2$. This is the detail that separates the hyperbola from the ellipse, where the relationship is $c^2 = a^2 - b^2$ instead — and getting these two confused is the most common error in conic-section problems.

The steps to find the foci, in order:

1. Write the equation in standard form so the right side equals $1$.

2. Read off $a^2$ (under the positive term) and $b^2$ (under the negative term).

3. Compute $c = \sqrt{a^2 + b^2}$.

4. Place the foci at $(\pm c, 0)$ for a horizontal hyperbola, or $(0, \pm c)$ for a vertical one.

How Are The Foci Related To Eccentricity?

The foci also set the hyperbola's eccentricity, a number measuring how "open" the curve is:

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

Since $c > a$ for every hyperbola, the eccentricity is always greater than $1$. A value just above $1$ gives a narrow, nearly-closed pair of branches; a large eccentricity gives wide, flat-opening branches. The foci can also be written as $(\pm ae, 0)$, which is handy when a problem gives you $a$ and $e$ directly.

Examples of Foci Of Hyperbola

The examples progress from reading off a standard equation to working back from eccentricity. Each step sits on its own line.

Example 1

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

Read off the values:

$$a^2 = 16, \quad b^2 = 9$$

Apply the focus relationship:

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

The hyperbola is horizontal, so the foci are at $(\pm c, 0)$.

Final answer: foci at $(5, 0)$ and $(-5, 0)$.

Example 2

Find the foci of $\dfrac{x^2}{36} - \dfrac{y^2}{64} = 1$. A student writes $c^2 = 36 - 64 = -28$ and says the foci do not exist. What went wrong?

The first instinct is to subtract — borrowing the relationship $b^2 = a^2 - c^2$ from the ellipse. Test where that leads: it produces $c^2 = -28$, a negative square, which is impossible. A negative result here is the tell-tale sign that the wrong relationship was used, because a hyperbola always has real foci.

The hyperbola uses addition:

$$c^2 = a^2 + b^2$$ $$c^2 = 36 + 64 = 100$$ $$c = 10$$

Final answer: foci at $(10, 0)$ and $(-10, 0)$. For a hyperbola, $c^2 = a^2 + b^2$ — never the ellipse's subtraction.

Example 3

Find the foci of the vertical hyperbola $\dfrac{y^2}{25} - \dfrac{x^2}{144} = 1$.

The positive term is under $y^2$, so the hyperbola opens vertically:

$$a^2 = 25, \quad b^2 = 144$$ $$c^2 = a^2 + b^2 = 25 + 144 = 169$$ $$c = 13$$

The foci lie on the y-axis at $(0, \pm c)$.

Final answer: foci at $(0, 13)$ and $(0, -13)$.

Example 4

A horizontal hyperbola has $a = 6$ and eccentricity $e = \dfrac{5}{3}$. Find its foci.

Use $c = ae$:

$$c = a e$$ $$c = 6 \times \frac{5}{3}$$ $$c = 10$$

Final answer: foci at $(10, 0)$ and $(-10, 0)$.

Example 5

A horizontal hyperbola has foci at $(\pm 13, 0)$ and a vertex at $(5, 0)$. Find $b^2$ and write its equation.

From the given points, $c = 13$ and $a = 5$. Solve for $b^2$:

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

The equation is therefore:

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

Final answer: $b^2 = 144$, equation $\dfrac{x^2}{25} - \dfrac{y^2}{144} = 1$.

Example 6

A hyperbola centered at $(2, -1)$ has $a = 3$ and $b = 4$, opening horizontally. Find the coordinates of its foci.

First find $c$:

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

For a center at $(h, k) = (2, -1)$, the horizontal foci sit at $(h \pm c, k)$:

$$(2 + 5, -1) = (7, -1)$$ $$(2 - 5, -1) = (-3, -1)$$

Final answer: foci at $(7, -1)$ and $(-3, -1)$.

Where The Foci Of A Hyperbola Do Real Work

The difference-of-distances property is not decorative — it is exactly what makes hyperbolas useful for finding position. In the LORAN navigation system, and in modern multilateration used to locate a phone or aircraft from signal timing, each pair of stations places the receiver on a hyperbola with those stations at the foci. Cross two such hyperbolas and the intersection pins down the location.

The foci also govern reflection. A signal aimed at one focus of a hyperbolic mirror reflects as though it came from the other focus — the principle behind the secondary mirror in a Cassegrain telescope, which folds a long light path into a short tube. The open shape of a hyperbola also describes the path of an object moving too fast to be captured by gravity, such as an interstellar comet swinging once past the Sun, and all of this rests on the two focal points and the constant $c = \sqrt{a^2 + b^2}$.

The hyperbola sits alongside the parabola and ellipse in the family of conic sections; compare its focus behaviour with the single focus of a parabola and the directrix of a parabola.

Common Mistakes With The Foci Of A Hyperbola

Mistake 1: Subtracting instead of adding under the square root

Where it slips in: Right after recalling the conic relationships, when the ellipse and hyperbola formulas blur together.

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

The correct way: A hyperbola uses $c^2 = a^2 + b^2$; only the ellipse subtracts. The first-instinct error here is importing the ellipse's subtraction — and it shows up as a negative number under the root, which can never happen for a real hyperbola. If you see a negative, you used the wrong relationship.

Mistake 2: Misreading which axis the hyperbola opens along

Where it slips in: Deciding whether the foci are on the x-axis or the y-axis.

Don't do this: Assume the foci are always horizontal because the example you remember was.

The correct way: The positive term sets the direction. If the $x^2$ term is positive, the hyperbola opens left-right and the foci are at $(\pm c, 0)$; if the $y^2$ term is positive, it opens up-down and the foci are at $(0, \pm c)$. The student who memorizes one diagram and applies it to every problem misplaces the foci by ninety degrees.

Mistake 3: Forgetting to shift for a non-origin center

Where it slips in: Hyperbolas centered at $(h, k)$ rather than the origin.

Don't do this: Report the foci as $(\pm c, 0)$ when the center is not at the origin.

The correct way: Add the center coordinates: the foci are at $(h \pm c, k)$ for a horizontal hyperbola. Compute $c$ from the axes first, then shift to the actual center.

Conclusion

• The foci of a hyperbola are two fixed points where the difference of distances to any curve point is the constant $2a$.

• They are found from $c^2 = a^2 + b^2$, giving foci at $(\pm c, 0)$ for a horizontal hyperbola.

• Eccentricity $e = c/a$ is always greater than $1$.

• The hyperbola adds ($c^2 = a^2 + b^2$) where the ellipse subtracts ($c^2 = a^2 - b^2$).

• For a center at $(h, k)$, shift the foci to $(h \pm c, k)$.

Practice And A Next Step

Work through the exercises above, and for each one identify the center and the opening direction before computing $c$. If a square comes out negative, that is your signal you subtracted where you should have added.

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