10.2Exploring Hyperbolas
Write hyperbola equations x²/a² − y²/b² = 1 in standard form. Identify vertices, foci (c² = a² + b²), and asymptotes y = ±(b/a)x. Graph and write equations from given information.
Hyperbolas model the paths of comets, the shape of cooling towers, and the LORAN navigation system. Their asymptotes make them unique among conics and introduce the concept of end behavior in a geometric context.
Essential Question
How does the hyperbola differ from the ellipse in its definition and equation, and what role do the asymptotes play in graphing?
Lesson Overview
Definition
A hyperbola is the set of all points where the DIFFERENCE of distances to two fixed points (foci) is constant and equal to 2a: |PF₁ − PF₂| = 2a.
Standard Forms
Horizontal (opens left/right)
x²/a² − y²/b² = 1
Foci at (±c, 0) · Asymptotes: y = ±(b/a)x · c² = a² + b²
Vertical (opens up/down)
y²/a² − x²/b² = 1
Foci at (0, ±c) · Asymptotes: y = ±(a/b)x · c² = a² + b²
Centered at (h, k)
(x−h)²/a² − (y−k)²/b² = 1
(y−k)²/a² − (x−h)²/b² = 1
⚠️ Critical Difference from Ellipse
For hyperbolas: c² = a² + b² (NOT a² − b²). The foci are outside the vertices, so c > a.
Key Features
- • Center: (h, k) — midpoint between foci and vertices
- • Vertices: on the transverse axis, distance a from center
- • Foci: distance c from center; c² = a² + b²
- • Asymptotes: lines the hyperbola approaches but never touches
- • Eccentricity: e = c/a > 1 (always greater than 1)
- • Asymptote box method: draw rectangle 2a × 2b centered at center; diagonals are asymptotes
Worked Examples
Identify all features of x²/9 − y²/16 = 1.
a² = 9 → a = 3; b² = 16 → b = 4
c² = 9 + 16 = 25 → c = 5
Horizontal hyperbola (positive x² term)
Center (0,0); Vertices (±3, 0); Foci (±5, 0)
Asymptotes: y = ±(4/3)x
Write the equation of a hyperbola with vertices (±2, 0) and foci (±√13, 0).
Horizontal hyperbola; a = 2 → a² = 4; c = √13 → c² = 13
b² = c² − a² = 13 − 4 = 9
Equation: x²/4 − y²/9 = 1
Find all features of (x+1)²/4 − (y−3)²/9 = 1.
Center (−1, 3); a² = 4 → a = 2; b² = 9 → b = 3
c² = 4 + 9 = 13 → c = √13
Horizontal; Vertices: (−1 ± 2, 3) = (1, 3) and (−3, 3)
Foci: (−1 ± √13, 3)
Asymptotes: y − 3 = ±(3/2)(x + 1)
Identify all features of y²/25 − x²/144 = 1.
Vertical hyperbola (y² term is positive)
a² = 25 → a = 5; b² = 144 → b = 12
c² = 25 + 144 = 169 → c = 13
Vertices: (0, ±5); Foci: (0, ±13)
Asymptotes: y = ±(5/12)x
Convert 4x² − 9y² − 16x + 54y − 101 = 0 to standard form.
Group: (4x² − 16x) − (9y² − 54y) = 101
Factor: 4(x² − 4x) − 9(y² − 6y) = 101
Complete square: 4(x−2)² − 16 − 9(y−3)² + 81 = 101
4(x−2)² − 9(y−3)² = 36
Divide by 36: (x−2)²/9 − (y−3)²/4 = 1
Guided Practice
Identify all features of x²/36 − y²/13 = 1.
Hint: c² = a² + b² for hyperbolas (not minus!). Find a, b, c, then state vertices, foci, and asymptotes.
Write the equation of a hyperbola with vertices (0, ±4) and foci (0, ±5).
Hint: Vertical hyperbola (vertices on y-axis). Use b² = c² − a².
Find the asymptotes of (x−2)²/16 − (y+1)²/9 = 1.
Hint: Asymptotes pass through the center (h,k) with slopes ±b/a.
Convert 9y² − 4x² − 18y + 24x − 63 = 0 to standard form.
Hint: Group y and x terms separately. Factor out 9 and −4 before completing the square.
A hyperbola has asymptotes y = ±(3/2)x and passes through (2, 3). Find the equation.
Hint: Use the asymptote slopes to find b/a = 3/2. Then substitute the point to find a².
Key Vocabulary
Hyperbola
Set of all points where the absolute difference of distances to two foci equals 2a.
Transverse axis
The axis connecting the two vertices; length 2a.
Conjugate axis
The axis perpendicular to the transverse axis; length 2b.
Vertices
Endpoints of the transverse axis, distance a from center.
Foci
Two fixed points; c² = a² + b² (foci are outside the vertices).
Asymptotes
Lines the hyperbola approaches but never touches; y − k = ±(b/a)(x − h) for a horizontal hyperbola.
Eccentricity
e = c/a > 1 for all hyperbolas.
Standard form
x²/a² − y²/b² = 1 (horizontal) or y²/a² − x²/b² = 1 (vertical).
Quick Check Quiz
Interactive Practice — 5 Questions
For x²/25 − y²/144 = 1, the foci are at:
The asymptotes of x²/4 − y²/9 = 1 are:
Which equation represents a vertical hyperbola?
For a hyperbola, c² equals:
The eccentricity of a hyperbola is always:
Independent Practice
Independent Practice
Find all features of x²/49 − y²/32 = 1.
Write the equation: vertices (0, ±6), foci (0, ±10).
Find all features of (y+2)²/9 − (x−1)²/16 = 1.
Write the equation: center (3, −1), horizontal, a = 4, b = 3.
Convert 4x² − y² + 24x + 4y + 28 = 0 to standard form.
Common Mistakes
Using c² = a² − b² for hyperbolas (like ellipses)
For hyperbolas: c² = a² + b². The foci are OUTSIDE the vertices, so c > a.
Thinking the asymptotes touch or cross the hyperbola
Asymptotes are lines the hyperbola approaches but NEVER touches. They are guides for graphing, not part of the curve.
Confusing which term is a² in a vertical hyperbola
In y²/a² − x²/b² = 1, a² is under y² (the positive term). The vertices are on the y-axis at (0, ±a).
Writing asymptote slopes as ±a/b for a horizontal hyperbola
For x²/a² − y²/b² = 1, asymptotes are y = ±(b/a)x — b is in the numerator, a in the denominator.
Math Tips
The key difference: ellipse has + between terms (x²/a² + y²/b² = 1); hyperbola has − (x²/a² − y²/b² = 1).
Asymptote box trick: draw a rectangle with width 2a and height 2b centered at (h,k). The diagonals of this box are the asymptotes.
For hyperbolas c² = a² + b², so c is always larger than both a and b. The foci are always outside the vertices.
To identify orientation: positive x² term → opens left/right; positive y² term → opens up/down.
Eccentricity e > 1 for hyperbolas. The larger e is, the more "open" (flat) the hyperbola looks.