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

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

TL;DR

The foci of a hyperbola are two fixed points whose difference of distances to any point on the curve is constant. They are found from (c^2 = a^2 + b^2), sitting at ((\pm c, 0)) for a horizontal hyperbola centered at the origin. This article covers the focus formula, the defining difference-of-distances property, eccentricity, and worked examples on the foci of a hyperbola — note this differs from the ellipse, where (c^2 = a^2 - b^2).

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, [ \frac{y^2}{a^2} - \frac{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.

Symbol Meaning
a Distance from center to a vertex (semi-transverse axis)
b Semi-conjugate axis (sets the asymptote slope)
c Distance from center to a focus
e Eccentricity, (e = \frac{c}{a}) (always > 1 for a hyperbola)

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), 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

Example 1

Find the foci of the hyperbola (\frac{x^2}{16} - \frac{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 ]

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

Example 2

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

The mistake is using the ellipse relationship:

[ b^2 = a^2 - c^2 ]

For hyperbolas, use addition:

[ c^2 = a^2 + b^2 ]

[ c^2 = 36 + 64 = 100 ]

[ c = 10 ]

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

Example 3

Find the foci of the vertical hyperbola (\frac{y^2}{25} - \frac{x^2}{144} = 1).

Read off the values:

(a^2 = 25, \quad b^2 = 144)

[ c^2 = 25 + 144 = 169 ]

[ c = 13 ]

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

Example 4

A horizontal hyperbola has (a = 6) and eccentricity (e = \frac{5}{3}). Find its foci.

Use (c = ae):

[ c = 6 \times \frac{5}{3} = 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 points, (c = 13) and (a = 5). Solve for (b^2):

[ c^2 = a^2 + b^2 \implies 169 = 25 + b^2 \implies b^2 = 144 ]

The equation is (\frac{x^2}{25} - \frac{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.

Find (c):

[ c^2 = a^2 + b^2 = 9 + 16 = 25 \implies c = 5 ]

The foci at ((h \pm c, k) = (2 \pm 5, -1)) are ((7, -1)) and ((-3, -1)).

Conclusion