Hyperbola Calculator

Enter the hyperbola equation or the known data to get its equations (standard and general form), its elements (center, foci, vertices, semi-axes, asymptotes, eccentricity, etc.), and its graph.

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Asymptote equation

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Asymptote equation

Quick Examples

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Solved Exercises

Determine the general form and elements of the hyperbola $ \dfrac{x^2}{16}-\dfrac{y^2}{9}=1. $

Hyperbola equations

Equation in standard form

$$ \dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 $$

Equation in general form

$$ 9x^2 - 16y^2 - 144 = 0 $$

Hyperbola elements

Orientation: Horizontal (transverse axis parallel to the x-axis, branches opening left and right).

Center: $ C \left(0, 0\right) $

Foci

$$ \begin{array}{l} F_1 \left(-5, 0\right) \\\\ F_2 \left(5, 0\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(-4, 0\right) \\\\ V_2 \left(4, 0\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(0, -3\right) \\\\ B_2 \left(0, 3\right) \end{array} $$

Equations of the asymptotes

$$ y = \pm \dfrac{3}{4} x $$

Semi-transverse axis: $ a = 4 $

Semi-conjugate axis: $ b = 3 $

Focal distance: $ c = 5 $

Latus rectum: $ L_R = \dfrac{9}{2} = 4.5 $

Eccentricity: $ e = \dfrac{5}{4} = 1.25 $

Axes of symmetry: $ x = 0, \quad y = 0 $

x-intercepts

$$ x_1 = -4 \\\\ x_2 = 4 $$

y-intercepts

There are no real intercepts.

Graph on the Cartesian plane of a horizontal hyperbola centered at the origin, example 1. — Graph of the hyperbola

Calculate the elements of the hyperbola with center not at the origin $ \dfrac{(y+3)^2}{10}-\dfrac{(x-2)^2}{20}=1. $

Equations

Equation in standard form

$$ \dfrac{\left(y + 3\right)^2}{10} - \dfrac{\left(x - 2\right)^2}{20} = 1 $$

Equation in general form

$$ x^2 - 2y^2 - 4x - 12y + 6 = 0 $$

Hyperbola elements

Orientation: Vertical (transverse axis parallel to the y-axis, branches opening up and down).

Center: $ C \left(2, -3\right) $

Foci

$$ \begin{array}{l} F_1 \left(2, -\sqrt{30}-3\right) \approx \left(2, -8.48\right) \\\\ F_2 \left(2, \sqrt{30}-3\right) \approx \left(2, 2.48\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(2, -\sqrt{10}-3\right) \approx \left(2, -6.16\right) \\\\ V_2 \left(2, \sqrt{10}-3\right) \approx \left(2, 0.16\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(-2 \sqrt{5}+2, -3\right) \approx \left(-2.47, -3\right) \\\\ B_2 \left(2 \sqrt{5}+2, -3\right) \approx \left(6.47, -3\right) \end{array} $$

Equations of the asymptotes

$$ y + 3 = \pm \dfrac{\sqrt{10}}{2 \sqrt{5}} \left(x - 2\right) $$

Semi-transverse axis: $ a = \sqrt{10} \approx 3.16 $

Semi-conjugate axis: $ b = 2 \sqrt{5} \approx 4.47 $

Focal distance: $ c = \sqrt{30} \approx 5.48 $

Latus rectum: $ L_R = \dfrac{40}{\sqrt{10}} \approx 12.65 $

Eccentricity: $ e = \dfrac{\sqrt{30}}{\sqrt{10}} \approx 1.73 $

Axes of symmetry: $ x = 2, \quad y = -3 $

x-intercepts

There are no real intercepts.

y-intercepts

$$ y_1 = 5\left(-\dfrac{2 \sqrt{3}}{5}-\dfrac{3}{5}\right) \approx -6.46 \\\\ y_2 = 5\left(\dfrac{2 \sqrt{3}}{5}-\dfrac{3}{5}\right) \approx 0.46 $$

Graph on the Cartesian plane of a vertical hyperbola centered away from the origin, example 2. — Graph of the hyperbola on the plane

Obtain the standard form and the elements of the hyperbola $ 4x^2-y^2=16. $

Equations

Equation in standard form

$$ \dfrac{x^2}{4} - \dfrac{y^2}{16} = 1 $$

Equation in general form

$$ 4x^2 - y^2 - 16 = 0 $$

Elements

Orientation: Horizontal (transverse axis parallel to the x-axis, branches opening left and right).

Center: $ C \left(0, 0\right) $

Foci

$$ \begin{array}{l} F_1 \left(-2 \sqrt{5}, 0\right) \approx \left(-4.47, 0\right) \\\\ F_2 \left(2 \sqrt{5}, 0\right) \approx \left(4.47, 0\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(-2, 0\right) \\\\ V_2 \left(2, 0\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(0, -4\right) \\\\ B_2 \left(0, 4\right) \end{array} $$

Asymptotes

$$ y = \pm 2 x $$

Semi-transverse axis: $ a = 2 $

Semi-conjugate axis: $ b = 4 $

Focal distance: $ c = 2 \sqrt{5} \approx 4.47 $

Latus rectum: $ L_R = 16 $

Eccentricity: $ e = \sqrt{5} \approx 2.24 $

Axes of symmetry: $ x = 0, \quad y = 0 $

x-intercepts

$$ x_1 = -2 \\\\ x_2 = 2 $$

y-intercepts

There are no real intercepts.

Graph on the Cartesian plane of a horizontal hyperbola centered at the origin, example 3. — Graph of the hyperbola on the Cartesian plane

Calculate the equation in standard form of the hyperbola $ 2y^2-x^2+2x+8y+3=0 $ and its elements.

Hyperbola equations

Equation in standard form

$$ \dfrac{\left(y + 2\right)^2}{2} - \dfrac{\left(x - 1\right)^2}{4} = 1 $$

Equation in general form

$$ x^2 - 2y^2 - 2x - 8y - 3 = 0 $$

Elements

Orientation: Vertical (transverse axis parallel to the y-axis, branches opening up and down).

Center: $ C \left(1, -2\right) $

Foci

$$ \begin{array}{l} F_1 \left(1, -\sqrt{6}-2\right) \approx \left(1, -4.45\right) \\\\ F_2 \left(1, \sqrt{6}-2\right) \approx \left(1, 0.45\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(1, -\sqrt{2}-2\right) \approx \left(1, -3.41\right) \\\\ V_2 \left(1, \sqrt{2}-2\right) \approx \left(1, -0.59\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(-1, -2\right) \\\\ B_2 \left(3, -2\right) \end{array} $$

Equations of the asymptotes

$$ y + 2 = \pm \dfrac{\sqrt{2}}{2} \left(x - 1\right) $$

Semi-transverse axis: $ a = \sqrt{2} \approx 1.41 $

Semi-conjugate axis: $ b = 2 $

Focal distance: $ c = \sqrt{6} \approx 2.45 $

Latus rectum: $ L_R = \dfrac{8}{\sqrt{2}} \approx 5.66 $

Eccentricity: $ e = \dfrac{\sqrt{6}}{\sqrt{2}} \approx 1.73 $

Axes of symmetry: $ x = 1, \quad y = -2 $

x-intercepts

$$ x_1 = 3 \\\\ x_2 = -1 $$

y-intercepts

$$ y_1 = \dfrac{-2 \sqrt{10}-8}{4} \approx -3.58 \\\\ y_2 = \dfrac{2 \sqrt{10}-8}{4} \approx -0.42 $$

Graph on the Cartesian plane of a vertical hyperbola centered away from the origin, example 4. — Graph of the hyperbola

Find the equations and elements of the hyperbola with foci at (±5, 0) and a vertex at (3, 0).

Equations

Equation in standard form

$$ \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1 $$

Equation in general form

$$ 16x^2 - 9y^2 - 144 = 0 $$

Hyperbola elements

Orientation: Horizontal (transverse axis parallel to the x-axis, branches opening left and right).

Center: $ C \left(0, 0\right) $

Foci

$$ \begin{array}{l} F_1 \left(-5, 0\right) \\\\ F_2 \left(5, 0\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(-3, 0\right) \\\\ V_2 \left(3, 0\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(0, -4\right) \\\\ B_2 \left(0, 4\right) \end{array} $$

Asymptotes

$$ y = \pm \dfrac{4}{3} x $$

Semi-transverse axis: $ a = 3 $

Semi-conjugate axis: $ b = 4 $

Focal distance: $ c = 5 $

Latus rectum: $ L_R = \dfrac{32}{3} \approx 10.67 $

Eccentricity: $ e = \dfrac{5}{3} \approx 1.67 $

Axes of symmetry: $ x = 0, \quad y = 0 $

x-intercepts

$$ x_1 = -3 \\\\ x_2 = 3 $$

y-intercepts

There are no real intercepts.

Graph on the Cartesian plane of a horizontal hyperbola centered at the origin, example 5. — Graph of the hyperbola

Determine the equations and elements of the hyperbola with vertices at (0, ±12) and asymptote y = -2x.

Equations

Equation in standard form

$$ \dfrac{y^2}{144} - \dfrac{x^2}{36} = 1 $$

Equation in general form

$$ 4x^2 - y^2 + 144 = 0 $$

Elements

Orientation: Vertical (transverse axis parallel to the y-axis, branches opening up and down).

Center: $ C \left(0, 0\right) $

Foci

$$ \begin{array}{l} F_1 \left(0, -6 \sqrt{5}\right) \approx \left(0, -13.42\right) \\\\ F_2 \left(0, 6 \sqrt{5}\right) \approx \left(0, 13.42\right)\end{array} $$

Vertices (endpoints of the transverse axis)

$$ \begin{array}{l} V_1 \left(0, -12\right) \\\\ V_2 \left(0, 12\right) \end{array} $$

Co-vertices (endpoints of the conjugate axis)

$$ \begin{array}{l} B_1 \left(-6, 0\right) \\\\ B_2 \left(6, 0\right) \end{array} $$

Equation of the asymptotes

$$ y = \pm 2 x $$

Semi-transverse axis: $ a = 12 $

Semi-conjugate axis: $ b = 6 $

Focal distance: $ c = 6 \sqrt{5} \approx 13.42 $

Latus rectum: $ L_R = 6 $

Eccentricity: $ e = \dfrac{\sqrt{5}}{2} \approx 1.12 $

Axes of symmetry: $ x = 0, \quad y = 0 $

x-intercepts

There are no real intercepts.

y-intercepts

$$ y_1 = -12 \\\\ y_2 = 12 $$

Graph on the Cartesian plane of a vertical hyperbola centered at the origin, example 6. — Graph of the hyperbola

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Daniel Machado

Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.