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Last modified on August 31st, 2024

The foci (singular: focus) of a hyperbola are two fixed points on its major axis, equidistant from the center. For any point on the hyperbola, the absolute difference between its distances to the two foci remains constant.

For a hyperbola given by the equation ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$, the foci are located at the coordinates (-c, 0) and (c, 0). Here, *c* is the distance between the center and the focus.

Thus, a graph of a hyperbola has two foci that are equidistant from the center.

Similarly, for any point on the hyperbola, the ratio of its distance from the foci to its distance from the directrix is always a constant called eccentricity, which is always greater than 1 (e > 1). Mathematically, ${e=\dfrac{c}{a}}$, thus the coordinates of the foci are (-ae, 0) and (ae, 0).

Also, the midpoint of the foci is the center of the hyperbola

The foci, center (midpoint of foci), and vertices of a hyperbola are collinear and lie on the same axis.

Here are the formulas to calculate the foci for the different types of hyperbolas.

Hyperbola | Foci |
---|---|

${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$ | (±c, 0), where c^{2} = a^{2} + b^{2} |

${\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1}$ | (0, ±c), where c^{2} = a^{2} + b^{2} |

${\dfrac{\left( x-h\right) ^{2}}{a^{2}}-\dfrac{\left( y-k\right) ^{2}}{b^{2}}=1}$ | (h ± c, k), where c^{2} = a^{2} + b^{2} |

${\dfrac{\left( y-k\right) ^{2}}{a^{2}}-\dfrac{\left( x-h\right) ^{2}}{b^{2}}=1}$ | (h, k ± c), where c^{2} = a^{2} + b^{2} |

The first step to finding the foci of a hyperbola is to identify whether the hyperbola is horizontal or vertical from its equation.

Let us find the foci for the hyperbola ${\dfrac{x^{2}}{36}-\dfrac{y^{2}}{9}=1}$

**Identifying the Hyperbola**

Since the given hyperbola** **${\dfrac{x^{2}}{36}-\dfrac{y^{2}}{9}=1}$ is in the form ${\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1}$, it is a horizontal hyperbola.

**Determining the Center of the Hyperbola and the Integers a and b**

Comparing with the standard equation, we get

- The center of the hyperbola is at (0, 0)
- a = 6 and b = 3

**Using the Formula c**^{2}** = a**^{2}** + b**^{2}** **

c^{2} = 6^{2} + 3^{2}

⇒ c^{2} = 45

⇒ c = ${\sqrt{45}}$

**Finding the Coordinates of the Foci **

As we know, the foci of the hyperbola are (-c, 0) and (c, 0)

Thus, its coordinates are ${\left( -\sqrt{45},0\right)}$ and ${\left( \sqrt{45},0\right)}$

**Problem – **Finding the Foci of the Hyperbola given by the equation **${\dfrac{\left( x-h\right) ^{2}}{a^{2}}-\dfrac{\left( y-k\right) ^{2}}{b^{2}}=1}$ **

**Find the foci of the hyperbola given by the equation****${\dfrac{\left( x+5\right) ^{2}}{81}-\dfrac{\left( y-7\right) ^{2}}{144}=1}$ **

Solution:

Since the given equation ${\dfrac{\left( x+5\right) ^{2}}{81}-\dfrac{\left( y-7\right) ^{2}}{144}=1}$ …..(i)

is in the form ${\dfrac{\left( x-h\right) ^{2}}{a^{2}}-\dfrac{\left( y-k\right) ^{2}}{b^{2}}=1}$ …..(ii), it is a horizontal hyperbola

On Comparing (i) and (ii), we get

The center of the hyperbola is at (h, k) = (-5, 7)

a = 9 and b = 12

Using the Formula c^{2} = a^{2} + b^{2}

c^{2} = 9^{2} + 12^{2}

⇒ c^{2} = 225

⇒ c = ${\sqrt{225}}$ = 15

Now, the foci of the hyperbola are (h – c, k) and (h + c, k)

⇒ (-5 – 15, 7) and (-5 + 15, 7)

Thus, the coordinates of the foci are (-20, 7) and (10, 7)

Now, if we are given to find the foci of the hyperbola ${\dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1}$, we will follow the following steps:

**Identifying the Hyperbola**

Since the given hyperbola** **${\dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1}$ is similar to its standard form ${\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1}$, it is a vertical hyperbola.

**Determining the Center of the Hyperbola and the Integers a and b**

Comparing with the standard equation, we get

- The center of the hyperbola is at (0, 0)
- a = 5 and b = 12

**Using the Formula c**^{2}** = a**^{2}** + b**^{2}** **

c^{2} = 5^{2} + 12^{2}

⇒ c^{2} = 169

⇒ c = ${\sqrt{169}}$ = 13

**Finding the Coordinates of the Foci **

As we know, the foci of the hyperbola are (0, c) and (0, -c)

Thus, its coordinates are (0, 13) and (0, -13)

**Problem – **Finding the Foci of the Hyperbola given by the equation **${\dfrac{\left( y-k\right) ^{2}}{a^{2}}-\dfrac{\left( x-h\right) ^{2}}{b^{2}}=1}$**

**Find the foci of the hyperbola given by the equation****${\dfrac{\left( y-2\right) ^{2}}{16}-\dfrac{\left( x+3\right) ^{2}}{64}=1}$**

Solution:

Since the given hyperbola ${\dfrac{\left( y-2\right) ^{2}}{16}-\dfrac{\left( x+3\right) ^{2}}{64}=1}$ …..(i)

is in the form ${\dfrac{\left( y-k\right) ^{2}}{a^{2}}-\dfrac{\left( x-h\right) ^{2}}{b^{2}}=1}$ …..(ii) it is a vertical hyperbola.

OnComparing (i) and (ii), we get

The center of the hyperbola is at (h, k) = (2, -3)

a = 4 and b = 8

Using the Formula c^{2} = a^{2} + b^{2}

c^{2} = 4^{2} + 8^{2}

⇒ c^{2} = 80

⇒ c = ${\sqrt{80}}$ = ${4\sqrt{5}}$

As we know, the foci of the hyperbola are (h, k + c) and (h, k – c)

⇒ ${\left( 2,-3+4\sqrt{5}\right)}$ and ${\left( 2,-3-4\sqrt{5}\right)}$

Thus, the coordinates of the foci are ${\left( 2,-3+4\sqrt{5}\right)}$ and ${\left( 2,-3-4\sqrt{5}\right)}$

Last modified on August 31st, 2024