AP PRECALCULUS — UNIT 4A · FUNCTIONS INVOLVING PARAMETERS, VECTORS, AND MATRICES
4.6C — Conic Sections — Hyperbolas
Notes — Two Branches and Asymptotes
💡 Learning Objectives (4.6.A Part 3)
By the end of this lesson you will be able to:
- Define a hyperbola in terms of two foci and a difference of distances
- Identify center, vertices, foci, and asymptotes from the equation
- Write the standard form equation of a hyperbola from given features
- Sketch a hyperbola including the central rectangle and asymptotes
1. Definition of a Hyperbola
A hyperbola is the set of all points whose distances to two fixed points (foci) DIFFER by a constant.
Where an ellipse uses the SUM of distances, the hyperbola uses the DIFFERENCE. The constant is 2a, where a is the distance from center to vertex.
2. Standard Forms
💡 Standard Forms of a Hyperbola Centered at (h, k)
- Horizontal transverse axis: ((x − h)²)/a² − ((y − k)²)/b² = 1
- Vertical transverse axis: ((y − k)²)/a² − ((x − h)²)/b² = 1
- a = distance from center to vertex (along transverse axis)
- Foci at distance c from center, where c² = a² + b² (different sign than ellipse!)
⚠️ How to tell horizontal vs. vertical transverse axis
For a hyperbola, look at which TERM IS POSITIVE. If the (x − h)² term is positive, the transverse axis is horizontal (the curve opens left-right). If the (y − k)² term is positive, the transverse axis is vertical (opens up-down). Unlike ellipses, the SIZE of the denominators doesn't tell direction.
3. Asymptotes
Hyperbolas have ASYMPTOTES — lines that the branches approach as the curve extends to infinity. For a horizontal-axis hyperbola centered at (h, k):
y − k = ±(b/a)(x − h)
For a vertical-axis hyperbola, the slopes flip:
y − k = ±(a/b)(x − h)
4. The Central Rectangle
To sketch a hyperbola, first draw a rectangle of width 2a and height 2b (or 2b and 2a, depending on orientation), centered at (h, k). The asymptotes are the EXTENDED DIAGONALS of this rectangle.
5. Worked Example
📘 Example — Identify features of (x − 1)²/16 − (y + 2)²/9 = 1
- Center: (1, −2)
- a² = 16, so a = 4; b² = 9, so b = 3
- Horizontal transverse axis (since x-term is positive)
- Vertices: (1 ± 4, −2) = (5, −2) and (−3, −2)
- c² = a² + b² = 16 + 9 = 25, so c = 5
- Foci: (1 ± 5, −2) = (6, −2) and (−4, −2)
- Asymptotes: y + 2 = ±(3/4)(x − 1)
6. Writing the Equation from Features
📘 Example — Write the equation
Hyperbola with vertices (3, 1) and (−1, 1), foci (4, 1) and (−2, 1).
- Center: midpoint of vertices = (1, 1)
- a = distance center to vertex = 2; horizontal transverse axis (vertices share y)
- c = distance center to focus = 3
- b² = c² − a² = 9 − 4 = 5
- Equation: (x − 1)²/4 − (y − 1)²/5 = 1
7. Eccentricity
For a hyperbola, e = c/a as well, but here e > 1. Higher eccentricity means the branches are more open and steeper; e close to 1 means the branches are nearly perpendicular to the transverse axis.
8. Sketching a Hyperbola
- Plot center
- Mark a units along the transverse axis from center to find the vertices
- Mark b units along the conjugate axis (perpendicular) to find ‘tips’ of the central rectangle
- Draw the central rectangle (width 2a, height 2b for horizontal axis case)
- Extend the diagonals of the rectangle — these are the asymptotes
- Mark c units along the transverse axis to find foci
- Draw the two branches, each starting at a vertex and approaching the asymptotes
9. Converting from General Form
📘 Example — Convert 9x² − 4y² − 36x − 8y − 4 = 0
- Group: 9(x² − 4x) − 4(y² + 2y) = 4
- Complete the square: 9((x − 2)² − 4) − 4((y + 1)² − 1) = 4
- Expand: 9(x − 2)² − 36 − 4(y + 1)² + 4 = 4
- Combine: 9(x − 2)² − 4(y + 1)² = 36
- Divide by 36: (x − 2)²/4 − (y + 1)²/9 = 1
- Center (2, −1), a = 2, b = 3, horizontal transverse axis.
10. Summary
- Hyperbola: difference of distances to two foci is constant 2a
- Standard form: positive term identifies the transverse axis direction
- c² = a² + b² (different sign than for ellipses)
- Asymptotes: diagonals of the central rectangle
- Eccentricity e = c/a > 1 always for a hyperbola