How to Find a Hyperbola’s Equation

A hyperbola is an important conic section in mathematics, characterized by its two distinct branches. To find a hyperbola’s equation, we need to understand its basic components and properties.

Key Components of a Hyperbola

Foci and Vertices

  • Foci: These are two fixed points used to define the hyperbola. The distance between any point on the hyperbola and the foci is constant.
  • Vertices: These are the points where each branch of the hyperbola is closest to the center.

Axes

  • Transverse Axis: The line segment that passes through the center and the vertices.
  • Conjugate Axis: The line segment perpendicular to the transverse axis through the center.

Standard Form of a Hyperbola’s Equation

There are two standard forms of a hyperbola’s equation, depending on its orientation:

  1. Horizontal Hyperbola: The transverse axis is horizontal.

    The equation is:

    $frac{(x – h)^2}{a^2} – frac{(y – k)^2}{b^2} = 1$

    Here, $(h, k)$ is the center, $a$ is the distance from the center to a vertex along the x-axis, and $b$ is the distance from the center to a vertex along the y-axis.

  2. Vertical Hyperbola: The transverse axis is vertical.

    The equation is:

    $frac{(y – k)^2}{a^2} – frac{(x – h)^2}{b^2} = 1$

    Here, $(h, k)$ is the center, $a$ is the distance from the center to a vertex along the y-axis, and $b$ is the distance from the center to a vertex along the x-axis.

Steps to Find the Equation

  1. Identify the Center: Determine the coordinates of the center $(h, k)$
  2. Determine the Distances: Measure the distance $a$ from the center to the vertices and the distance $b$ from the center to the co-vertices.
  3. Select the Orientation: Decide if the hyperbola is horizontal or vertical.
  4. Plug into the Formula: Use the appropriate standard form equation mentioned above.

Example

Let’s find the equation of a hyperbola with a center at $(2, 3)$, vertices at $(5, 3)$ and $(-1, 3)$, and co-vertices at $(2, 6)$ and $(2, 0)$

  1. Identify the Center: The center is $(2, 3)$

  2. Determine the Distances:

    • Distance $a$ (from center to vertices) is $|5 – 2| = 3$
    • Distance $b$ (from center to co-vertices) is $|6 – 3| = 3$

  3. Select the Orientation: Since the vertices are along the horizontal axis, it is a horizontal hyperbola.

  4. Plug into the Formula:

    $frac{(x – 2)^2}{3^2} – frac{(y – 3)^2}{3^2} = 1$

    Simplifying, we get:

    $frac{(x – 2)^2}{9} – frac{(y – 3)^2}{9} = 1$

Conclusion

Understanding the components and standard forms of a hyperbola is crucial for deriving its equation. By following the steps and using the appropriate formula, you can easily find the equation of any hyperbola.

Citations

  1. 1. Khan Academy – Hyperbolas
  2. 2. Math is Fun – Hyperbola
  3. 3. Purplemath – Hyperbolas

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ