Mathematics
Grade 12
15 min
Write equations of hyperbolas in standard form
Write equations of hyperbolas in standard form
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1
Introduction & Learning Objectives
Learning Objectives
Identify the center, vertices, and foci of a hyperbola from given information.
Determine whether a hyperbola has a horizontal or vertical transverse axis.
Calculate the values of 'a', 'b', and 'c' for a hyperbola.
Apply the relationship c² = a² + b² to find the missing parameter for a hyperbola's equation.
Write the final equation of a hyperbola in standard form given its key characteristics.
Derive the equation of a hyperbola using information about its asymptotes.
Have you ever wondered about the shape of a nuclear cooling tower or the shockwave of a supersonic jet? 🏗️✈️ Those are real-world hyperbolas!
This tutorial will guide you through the process of translating the geometric properties of a hyperbola—like its cen...
2
Key Concepts & Vocabulary
TermDefinitionExample
Center (h, k)The midpoint of the line segment connecting the foci, and also the midpoint of the line segment connecting the vertices. It is the point of intersection of the hyperbola's axes of symmetry.For a hyperbola with vertices at (5, 2) and (-1, 2), the center is the midpoint, which is (2, 2).
Transverse AxisThe line segment that connects the two vertices of the hyperbola. Its orientation (horizontal or vertical) determines the standard form of the equation.If the vertices are at (3, 0) and (-3, 0), the transverse axis is horizontal and has a length of 6 units.
VerticesThe two points where the hyperbola intersects its transverse axis. The distance from the center to each vertex is denoted by 'a'.If the center is (0, 0) and a = 4 with a horizontal...
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Core Formulas
Standard Form (Horizontal Transverse Axis)
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1
Use this form when the vertices and foci lie on a horizontal line. The x-term is positive. The center is (h, k), 'a' is the distance from the center to a vertex, and 'b' is related to the conjugate axis.
Standard Form (Vertical Transverse Axis)
\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1
Use this form when the vertices and foci lie on a vertical line. The y-term is positive. The center is (h, k), 'a' is the distance from the center to a vertex, and 'b' is related to the conjugate axis.
Focal Relationship
c^2 = a^2 + b^2
This fundamental relationship connects the distances 'a' (center-to-vertex), 'b' (related to the con...
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Challenging
Find the standard form of the equation of the hyperbola with foci at (2, 9) and (2, -7) and asymptotes y = \pm \frac{3}{4}x + \frac{5}{2}.
A.\frac{(y-1)^2}{36} - \frac{(x-2)^2}{64} = 1
B.\frac{(x-2)^2}{64} - \frac{(y-1)^2}{36} = 1
C.\frac{(y-1)^2}{36} - \frac{(x-2)^2}{64} = 1
D.\frac{(y-1)^2}{64} - \frac{(x-2)^2}{36} = 1
Challenging
A hyperbola has vertices at (7, -2) and (-1, -2). Its asymptotes are given by the equations y+2 = \pm \frac{5}{4}(x-3). Find the equation of the hyperbola.
A.\frac{(y+2)^2}{16} - \frac{(x-3)^2}{25} = 1
B.\frac{(x-3)^2}{16} - \frac{(y+2)^2}{25} = 1
C.\frac{(x-3)^2}{25} - \frac{(y+2)^2}{16} = 1
D.\frac{(x-3)^2}{16} - \frac{(y+2)^2}{16} = 1
Challenging
Find the equation of the hyperbola with foci at (3, 6) and (3, -4) and endpoints of the conjugate axis at (0, 1) and (6, 1).
A.\frac{(y-1)^2}{16} - \frac{(x-3)^2}{9} = 1
B.\frac{(x-3)^2}{9} - \frac{(y-1)^2}{16} = 1
C.\frac{(y-1)^2}{9} - \frac{(x-3)^2}{16} = 1
D.\frac{(y-1)^2}{16} - \frac{(x-3)^2}{25} = 1
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