Lesson 4: Hyperbolas
Learning Objectives
- Understand the definition of a hyperbola using two foci and the difference of distances
- Identify and use the standard equations for horizontal and vertical hyperbolas
- Find the center, vertices, co-vertices, and foci of a hyperbola
- Calculate c using the relationship c² = a² + b²
- Write and graph equations of asymptotes for hyperbolas
- Graph hyperbolas using the central rectangle method
- Write equations of hyperbolas from given information
- Calculate and interpret eccentricity of hyperbolas
- Apply hyperbolas to real-world problems including navigation and physics
Introduction: What is a Hyperbola?
Hyperbola: The set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (called foci) is constant.
If P is any point on the hyperbola and F₁ and F₂ are the foci:
|d₁ - d₂| = constant
A hyperbola has two separate branches that open away from each other. Unlike an ellipse, where the sum of distances is constant, a hyperbola uses the difference of distances.
Key Components of a Hyperbola
- Center (h, k): The midpoint between the two foci
- Vertices: The two points on the hyperbola closest to the center
- Foci (plural of focus): Two fixed points used to define the hyperbola
- Transverse axis: The line segment connecting the vertices (length = 2a)
- Conjugate axis: The perpendicular axis through the center (length = 2b)
- Asymptotes: Two diagonal lines that the hyperbola approaches but never touches
Important Distinction:
- Ellipse: Sum of distances is constant → c² = a² - b²
- Hyperbola: Difference of distances is constant → c² = a² + b²
Notice the sign change! For hyperbolas, we ADD a² and b² to get c².
Section 1: Standard Equations of Hyperbolas
The standard form of a hyperbola's equation depends on whether it opens horizontally or vertically.
Horizontal Hyperbola (opens left and right)
Standard Equation:
(x - h)²/a² - (y - k)²/b² = 1
Characteristics of Horizontal Hyperbolas:
- Center: (h, k)
- Vertices: (h ± a, k) — located a units left and right from center
- Co-vertices: (h, k ± b) — used to construct the central rectangle
- Foci: (h ± c, k) where c² = a² + b²
- Asymptotes: y - k = ±(b/a)(x - h)
- Transverse axis is horizontal
Vertical Hyperbola (opens up and down)
Standard Equation:
(y - k)²/a² - (x - h)²/b² = 1
Characteristics of Vertical Hyperbolas:
- Center: (h, k)
- Vertices: (h, k ± a) — located a units up and down from center
- Co-vertices: (h ± b, k) — used to construct the central rectangle
- Foci: (h, k ± c) where c² = a² + b²
- Asymptotes: y - k = ±(a/b)(x - h)
- Transverse axis is vertical
How to Identify Horizontal vs. Vertical:
Look at which variable comes first (has the positive sign):
- If x² is positive → Horizontal hyperbola
- If y² is positive → Vertical hyperbola
Example 1: Identifying Horizontal vs. Vertical Hyperbolas
Determine whether each hyperbola is horizontal or vertical:
a) (x - 2)²/9 - (y + 1)²/16 = 1
b) (y - 3)²/25 - (x + 4)²/4 = 1
Solution:
Part a:
The x² term is positive, so this is a horizontal hyperbola.
Part b:
The y² term is positive, so this is a vertical hyperbola.
Answer: (a) Horizontal; (b) Vertical
Example 2: Finding the Center and Values of a and b
For the hyperbola (x - 3)²/16 - (y + 2)²/9 = 1, find the center, a, and b.
Solution:
Step 1: Identify the center (h, k) from the equation form (x - h)²/a² - (y - k)²/b² = 1.
(x - 3)² → h = 3
(y + 2)² = (y - (-2))² → k = -2
Center: (3, -2)
Step 2: Identify a² and b².
Since the x² term is positive, this is horizontal.
a² = 16 → a = 4
b² = 9 → b = 3
Answer: Center (3, -2); a = 4; b = 3
Example 3: Finding Vertices and Foci (Horizontal Hyperbola)
For the hyperbola (x + 1)²/25 - (y - 4)²/9 = 1, find the center, vertices, and foci.
Solution:
Step 1: Identify center, a, and b.
Center: (-1, 4)
a² = 25 → a = 5
b² = 9 → b = 3
This is a horizontal hyperbola (x² is positive).
Step 2: Find the vertices (h ± a, k).
Vertices: (-1 - 5, 4) and (-1 + 5, 4)
Vertices: (-6, 4) and (4, 4)
Step 3: Calculate c using c² = a² + b².
c² = 25 + 9 = 34
c = √34 ≈ 5.83
Step 4: Find the foci (h ± c, k).
Foci: (-1 - √34, 4) and (-1 + √34, 4)
Foci: (-1 - 5.83, 4) and (-1 + 5.83, 4)
Foci: (-6.83, 4) and (4.83, 4)
Answer: Center (-1, 4); Vertices (-6, 4) and (4, 4); Foci (-1 ± √34, 4)
Example 4: Finding Vertices and Foci (Vertical Hyperbola)
For the hyperbola (y + 2)²/36 - (x - 1)²/16 = 1, find the center, vertices, and foci.
Solution:
Step 1: Identify center, a, and b.
Center: (1, -2)
a² = 36 → a = 6
b² = 16 → b = 4
This is a vertical hyperbola (y² is positive).
Step 2: Find the vertices (h, k ± a).
Vertices: (1, -2 - 6) and (1, -2 + 6)
Vertices: (1, -8) and (1, 4)
Step 3: Calculate c using c² = a² + b².
c² = 36 + 16 = 52
c = √52 = 2√13 ≈ 7.21
Step 4: Find the foci (h, k ± c).
Foci: (1, -2 - 2√13) and (1, -2 + 2√13)
Foci: (1, -2 - 7.21) and (1, -2 + 7.21)
Foci: (1, -9.21) and (1, 5.21)
Answer: Center (1, -2); Vertices (1, -8) and (1, 4); Foci (1, -2 ± 2√13)
Check Your Understanding
Question 1: For the hyperbola x²/49 - y²/25 = 1, what is the value of c (distance from center to focus)?
Answer: c = √74 ≈ 8.60
Explanation: Since a² = 49 and b² = 25, we use c² = a² + b² = 49 + 25 = 74, so c = √74.
Question 2: Is the hyperbola (y - 5)²/100 - (x + 3)²/64 = 1 horizontal or vertical?
Answer: Vertical
Explanation: The y² term is positive, so the hyperbola opens vertically (up and down).
Section 2: Asymptotes of Hyperbolas
One of the most distinctive features of hyperbolas is their asymptotes — diagonal lines that the hyperbola approaches but never crosses. Asymptotes help us graph hyperbolas accurately.
Asymptotes: Lines that the branches of a hyperbola approach as they extend toward infinity.
Asymptote Equations
For Horizontal Hyperbolas: (x - h)²/a² - (y - k)²/b² = 1
Asymptotes: y - k = ±(b/a)(x - h)
For Vertical Hyperbolas: (y - k)²/a² - (x - h)²/b² = 1
Asymptotes: y - k = ±(a/b)(x - h)
The Central Rectangle Method
To graph a hyperbola and its asymptotes:
- Plot the center (h, k)
- From the center, mark a units along the transverse axis (toward vertices)
- From the center, mark b units along the conjugate axis (perpendicular direction)
- Draw a rectangle using these four points
- Draw the asymptotes through opposite corners of the rectangle
- Sketch the hyperbola branches opening toward the vertices and approaching the asymptotes
Example 5: Finding Asymptote Equations (Horizontal)
Find the equations of the asymptotes for (x - 2)²/16 - (y + 3)²/9 = 1.
Solution:
Step 1: Identify the center, a, and b.
Center: (2, -3)
a² = 16 → a = 4
b² = 9 → b = 3
This is horizontal (x² is positive).
Step 2: Use the asymptote formula y - k = ±(b/a)(x - h).
y - (-3) = ±(3/4)(x - 2)
y + 3 = ±(3/4)(x - 2)
Step 3: Write both equations.
Asymptote 1: y + 3 = (3/4)(x - 2) → y = (3/4)x - 3/2 - 3 → y = (3/4)x - 9/2
Asymptote 2: y + 3 = -(3/4)(x - 2) → y = -(3/4)x + 3/2 - 3 → y = -(3/4)x - 3/2
Answer: y = (3/4)x - 9/2 and y = -(3/4)x - 3/2
Example 6: Finding Asymptote Equations (Vertical)
Find the equations of the asymptotes for (y - 1)²/25 - (x + 2)²/4 = 1.
Solution:
Step 1: Identify the center, a, and b.
Center: (-2, 1)
a² = 25 → a = 5
b² = 4 → b = 2
This is vertical (y² is positive).
Step 2: Use the asymptote formula y - k = ±(a/b)(x - h).
y - 1 = ±(5/2)(x - (-2))
y - 1 = ±(5/2)(x + 2)
Step 3: Write both equations.
Asymptote 1: y - 1 = (5/2)(x + 2) → y = (5/2)x + 5 + 1 → y = (5/2)x + 6
Asymptote 2: y - 1 = -(5/2)(x + 2) → y = -(5/2)x - 5 + 1 → y = -(5/2)x - 4
Answer: y = (5/2)x + 6 and y = -(5/2)x - 4
Example 7: Finding Asymptotes for a Centered Hyperbola
Find the asymptote equations for x²/9 - y²/16 = 1.
Solution:
Step 1: Identify the center, a, and b.
Center: (0, 0)
a² = 9 → a = 3
b² = 16 → b = 4
This is horizontal (x² is positive).
Step 2: Use the asymptote formula y - 0 = ±(b/a)(x - 0).
y = ±(4/3)x
Answer: y = (4/3)x and y = -(4/3)x
Example 8: Complete Analysis of a Hyperbola
For x²/36 - y²/64 = 1, find the center, vertices, foci, and asymptotes.
Solution:
Step 1: Identify center, a, and b.
Center: (0, 0)
a² = 36 → a = 6
b² = 64 → b = 8
Horizontal hyperbola
Step 2: Find vertices (±a, 0).
Vertices: (-6, 0) and (6, 0)
Step 3: Calculate c and find foci.
c² = a² + b² = 36 + 64 = 100
c = 10
Foci: (-10, 0) and (10, 0)
Step 4: Find asymptotes y = ±(b/a)x.
y = ±(8/6)x = ±(4/3)x
Asymptotes: y = (4/3)x and y = -(4/3)x
Answer:
- Center: (0, 0)
- Vertices: (-6, 0) and (6, 0)
- Foci: (-10, 0) and (10, 0)
- Asymptotes: y = (4/3)x and y = -(4/3)x
Check Your Understanding
Question 3: For the hyperbola y²/49 - x²/16 = 1, what is the slope of the asymptotes?
Answer: ±7/4
Explanation: For a vertical hyperbola, asymptotes are y = ±(a/b)x. Here a² = 49 (a = 7) and b² = 16 (b = 4), so slopes are ±7/4.
Question 4: What are the asymptotes for (x + 1)²/4 - (y - 2)²/9 = 1?
Answer: y - 2 = ±(3/2)(x + 1) or y = (3/2)x + 7/2 and y = -(3/2)x + 1/2
Explanation: Horizontal hyperbola with center (-1, 2), a = 2, b = 3. Asymptotes: y - 2 = ±(3/2)(x + 1).
Section 3: Writing Equations of Hyperbolas
Given information about a hyperbola, we can write its equation in standard form.
Example 9: Writing an Equation from Center and Vertices (Horizontal)
Write the equation of a hyperbola with center (0, 0), vertices at (±5, 0), and foci at (±13, 0).
Solution:
Step 1: Determine the orientation.
The vertices and foci lie on the x-axis, so this is a horizontal hyperbola.
Step 2: Find a from the vertices.
Distance from center to vertex is a = 5.
Step 3: Find c from the foci.
Distance from center to focus is c = 13.
Step 4: Use c² = a² + b² to find b.
13² = 5² + b²
169 = 25 + b²
b² = 144
b = 12
Step 5: Write the equation.
x²/25 - y²/144 = 1
Answer: x²/25 - y²/144 = 1
Example 10: Writing an Equation from Center and Vertices (Vertical)
Write the equation of a hyperbola with center (2, -3), vertices at (2, 1) and (2, -7), and foci at (2, 2) and (2, -8).
Solution:
Step 1: Determine the orientation.
The vertices have the same x-coordinate, so this is a vertical hyperbola.
Step 2: Identify h and k.
Center: (h, k) = (2, -3)
Step 3: Find a.
Distance from center (2, -3) to vertex (2, 1) is |-3 - 1| = 4.
So a = 4.
Step 4: Find c.
Distance from center (2, -3) to focus (2, 2) is |-3 - 2| = 5.
So c = 5.
Step 5: Use c² = a² + b² to find b.
5² = 4² + b²
25 = 16 + b²
b² = 9
b = 3
Step 6: Write the equation.
(y - (-3))²/16 - (x - 2)²/9 = 1
(y + 3)²/16 - (x - 2)²/9 = 1
Answer: (y + 3)²/16 - (x - 2)²/9 = 1
Example 11: Writing an Equation from Asymptotes and a Vertex
Write the equation of a hyperbola centered at the origin with asymptotes y = ±(3/2)x and vertex at (2, 0).
Solution:
Step 1: Determine the orientation.
The vertex is on the x-axis, so this is a horizontal hyperbola.
Step 2: Find a from the vertex.
The vertex is at (2, 0), so a = 2.
Step 3: Use the asymptote slope to find b.
For a horizontal hyperbola, asymptote slope = b/a.
b/a = 3/2
b/2 = 3/2
b = 3
Step 4: Write the equation.
x²/4 - y²/9 = 1
Answer: x²/4 - y²/9 = 1
Example 12: Writing an Equation from Foci and Asymptote Slope
A horizontal hyperbola centered at the origin has foci at (±5, 0) and asymptotes with slope ±4/3. Write its equation.
Solution:
Step 1: Identify c.
Foci at (±5, 0) means c = 5.
Step 2: Use asymptote slope to relate a and b.
For horizontal hyperbola: b/a = 4/3
So b = (4/3)a
Step 3: Use c² = a² + b².
5² = a² + ((4/3)a)²
25 = a² + (16/9)a²
25 = (9/9)a² + (16/9)a²
25 = (25/9)a²
a² = 9
a = 3
Step 4: Find b.
b = (4/3)(3) = 4
Step 5: Write the equation.
x²/9 - y²/16 = 1
Answer: x²/9 - y²/16 = 1
Check Your Understanding
Question 5: A vertical hyperbola has center (0, 0), vertices at (0, ±6), and c = 10. What is b²?
Answer: b² = 64
Explanation: a = 6, c = 10. Using c² = a² + b²: 100 = 36 + b², so b² = 64.
Question 6: What is the equation of a horizontal hyperbola with center (1, -2), a = 3, and b = 4?
Answer: (x - 1)²/9 - (y + 2)²/16 = 1
Explanation: Use (x - h)²/a² - (y - k)²/b² = 1 with (h, k) = (1, -2), a² = 9, b² = 16.
Section 4: Converting General Form to Standard Form
Sometimes hyperbolas are given in general form. We need to complete the square to convert to standard form.
Example 13: Converting to Standard Form
Write 9x² - 4y² - 36x - 8y - 4 = 0 in standard form.
Solution:
Step 1: Group x terms and y terms.
(9x² - 36x) - (4y² + 8y) = 4
Step 2: Factor out the coefficients of x² and y².
9(x² - 4x) - 4(y² + 2y) = 4
Step 3: Complete the square for both variables.
For x: (x² - 4x + 4) requires adding 4 inside, which is 9(4) = 36 on the right
For y: (y² + 2y + 1) requires adding 1 inside, which is 4(1) = 4 on the right
9(x² - 4x + 4) - 4(y² + 2y + 1) = 4 + 36 - 4
9(x - 2)² - 4(y + 1)² = 36
Step 4: Divide by 36 to get 1 on the right.
9(x - 2)²/36 - 4(y + 1)²/36 = 1
(x - 2)²/4 - (y + 1)²/9 = 1
Answer: (x - 2)²/4 - (y + 1)²/9 = 1
Center: (2, -1); Horizontal hyperbola; a = 2, b = 3
Example 14: Converting to Standard Form (Vertical)
Write 4y² - 9x² + 16y + 18x - 29 = 0 in standard form.
Solution:
Step 1: Rearrange: positive term first.
4y² + 16y - 9x² + 18x = 29
Step 2: Factor out coefficients.
4(y² + 4y) - 9(x² - 2x) = 29
Step 3: Complete the square.
For y: (y² + 4y + 4) requires adding 4 inside, which is 4(4) = 16 on the right
For x: (x² - 2x + 1) requires adding 1 inside, which is 9(1) = 9 on the right
4(y² + 4y + 4) - 9(x² - 2x + 1) = 29 + 16 - 9
4(y + 2)² - 9(x - 1)² = 36
Step 4: Divide by 36.
(y + 2)²/9 - (x - 1)²/4 = 1
Answer: (y + 2)²/9 - (x - 1)²/4 = 1
Center: (1, -2); Vertical hyperbola; a = 3, b = 2
Example 15: Complete Analysis After Converting
For x² - 4y² - 2x - 16y - 19 = 0, convert to standard form and find center, vertices, foci, and asymptotes.
Solution:
Step 1: Convert to standard form.
(x² - 2x) - 4(y² + 4y) = 19
(x² - 2x + 1) - 4(y² + 4y + 4) = 19 + 1 - 16
(x - 1)² - 4(y + 2)² = 4
(x - 1)²/4 - (y + 2)²/1 = 1
Step 2: Identify center, a, and b.
Center: (1, -2)
Horizontal hyperbola (x² is positive)
a² = 4 → a = 2
b² = 1 → b = 1
Step 3: Find vertices.
Vertices: (1 ± 2, -2) = (-1, -2) and (3, -2)
Step 4: Calculate c and find foci.
c² = a² + b² = 4 + 1 = 5
c = √5
Foci: (1 ± √5, -2)
Step 5: Find asymptotes.
y - (-2) = ±(1/2)(x - 1)
y + 2 = ±(1/2)(x - 1)
y = (1/2)x - 1/2 - 2 = (1/2)x - 5/2
y = -(1/2)x + 1/2 - 2 = -(1/2)x - 3/2
Answer:
- Standard form: (x - 1)²/4 - (y + 2)²/1 = 1
- Center: (1, -2)
- Vertices: (-1, -2) and (3, -2)
- Foci: (1 - √5, -2) and (1 + √5, -2)
- Asymptotes: y = (1/2)x - 5/2 and y = -(1/2)x - 3/2
Check Your Understanding
Question 7: What is the center of the hyperbola 16x² - 9y² + 64x + 18y - 89 = 0?
Answer: (-2, 1)
Explanation: Complete the square: 16(x + 2)² - 9(y - 1)² = 144, giving center (-2, 1).
Section 5: Eccentricity of Hyperbolas
Eccentricity (e): A measure of how "stretched" a hyperbola is.
e = c/a
Properties of Eccentricity
- For all hyperbolas: e > 1
- As e approaches 1, the hyperbola becomes narrower (branches closer together)
- As e increases, the hyperbola becomes wider (branches farther apart)
- Compare: Circle (e = 0), Ellipse (0 < e < 1), Parabola (e = 1), Hyperbola (e > 1)
Example 16: Finding Eccentricity
Find the eccentricity of the hyperbola x²/16 - y²/9 = 1.
Solution:
Step 1: Identify a and b.
a² = 16 → a = 4
b² = 9 → b = 3
Step 2: Calculate c.
c² = a² + b² = 16 + 9 = 25
c = 5
Step 3: Calculate eccentricity.
e = c/a = 5/4 = 1.25
Answer: e = 5/4 or 1.25
Example 17: Comparing Eccentricities
Compare the eccentricities of these two hyperbolas:
a) x²/4 - y²/5 = 1
b) x²/4 - y²/96 = 1
Solution:
Part a:
a = 2, b² = 5
c² = 4 + 5 = 9 → c = 3
e = 3/2 = 1.5
Part b:
a = 2, b² = 96
c² = 4 + 96 = 100 → c = 10
e = 10/2 = 5
Comparison:
Hyperbola (b) has eccentricity 5, much greater than hyperbola (a) with eccentricity 1.5.
This means hyperbola (b) is much wider, with branches farther apart.
Answer: e(a) = 1.5; e(b) = 5. Hyperbola (b) is much wider.
Example 18: Finding a Hyperbola Given Eccentricity
A hyperbola centered at the origin has a = 3 and eccentricity e = 2. Write its equation if it's horizontal.
Solution:
Step 1: Use e = c/a to find c.
2 = c/3
c = 6
Step 2: Use c² = a² + b² to find b.
6² = 3² + b²
36 = 9 + b²
b² = 27
Step 3: Write the equation.
x²/9 - y²/27 = 1
Answer: x²/9 - y²/27 = 1
Check Your Understanding
Question 8: What is the eccentricity of the hyperbola y²/25 - x²/144 = 1?
Answer: e = 13/5 = 2.6
Explanation: a = 5, b = 12, c² = 25 + 144 = 169, so c = 13. Thus e = 13/5 = 2.6.
Section 6: Comparison of Ellipses and Hyperbolas
Key Differences to Remember
| Property | Ellipse | Hyperbola |
|---|---|---|
| Definition | Sum of distances is constant | Difference of distances is constant |
| Relationship | c² = a² - b² | c² = a² + b² |
| Shape | Closed curve (oval) | Two separate branches |
| Eccentricity | 0 < e < 1 | e > 1 |
| Asymptotes | None | Two diagonal lines |
| Vertices | On the major axis (larger denominator) | On the transverse axis (positive term) |
| Horizontal Equation | (x - h)²/a² + (y - k)²/b² = 1 | (x - h)²/a² - (y - k)²/b² = 1 |
| Vertical Equation | (y - k)²/a² + (x - h)²/b² = 1 | (y - k)²/a² - (x - h)²/b² = 1 |
Example 19: Identifying Conic Type
Identify whether each equation represents an ellipse or hyperbola:
a) (x - 1)²/9 + (y + 2)²/16 = 1
b) (y - 3)²/25 - (x + 1)²/9 = 1
c) 4x² + 9y² = 36
d) 9x² - 16y² = 144
Solution:
Part a:
Both terms are positive (addition) → Ellipse
Part b:
One term is positive, one negative (subtraction) → Hyperbola
Part c:
Divide by 36: x²/9 + y²/4 = 1
Both positive → Ellipse
Part d:
Divide by 144: x²/16 - y²/9 = 1
Subtraction → Hyperbola
Answer: (a) Ellipse; (b) Hyperbola; (c) Ellipse; (d) Hyperbola
Section 7: Applications of Hyperbolas
Hyperbolas appear in many real-world applications, particularly in navigation, physics, and architecture.
Example 20: LORAN Navigation System
LORAN (Long Range Navigation) uses hyperbolas to determine a ship's position. Two radio stations are located at (-100, 0) and (100, 0) (in miles). A ship receives signals from both stations. If the difference in arrival times indicates the ship is 120 miles closer to the eastern station, find the equation of the hyperbola on which the ship is located.
Solution:
Step 1: Understand the setup.
The two stations are the foci at F₁(-100, 0) and F₂(100, 0).
The ship is on a hyperbola where the difference of distances is constant.
Step 2: Identify c.
Distance from center (0, 0) to focus is c = 100 miles.
Step 3: Identify 2a.
The difference in distances is 120 miles, so 2a = 120.
Therefore, a = 60 miles.
Step 4: Find b using c² = a² + b².
100² = 60² + b²
10,000 = 3,600 + b²
b² = 6,400
b = 80
Step 5: Write the equation.
Since the ship is closer to the eastern station (right), it's on the right branch.
x²/3600 - y²/6400 = 1
Answer: x²/3600 - y²/6400 = 1 (where x and y are in miles)
Example 21: Hyperbolic Mirror
A hyperbolic mirror has the property that light rays directed toward one focus reflect toward the other focus. A mirror has the equation x²/9 - y²/16 = 1 (in meters). If a light source is placed at one focus, where is the reflected focal point?
Solution:
Step 1: Identify a and b.
a² = 9 → a = 3 meters
b² = 16 → b = 4 meters
Step 2: Calculate c.
c² = a² + b² = 9 + 16 = 25
c = 5 meters
Step 3: Find the foci.
This is a horizontal hyperbola centered at the origin.
Foci: (-5, 0) and (5, 0)
Answer: If the light source is at (-5, 0), the reflected focal point is at (5, 0), or vice versa.
Example 22: Cooling Tower Design
A nuclear power plant cooling tower has a hyperbolic shape. The tower is 150 meters tall. At the base (ground level), the radius is 50 meters. At the narrowest point (75 meters high), the radius is 30 meters. Model the cross-section of the tower using a hyperbola with center at the narrowest point. What is the radius at the top (150 meters high)?
Solution:
Step 1: Set up coordinate system.
Place center at (0, 0) at the narrowest point.
The tower extends 75 meters up and 75 meters down.
At the narrowest point: radius = 30 meters, so a = 30.
Step 2: Use a point to find b.
At y = -75 (base), x = 50 (radius).
The equation is x²/900 - y²/b² = 1
Substitute (50, -75):
50²/900 - (-75)²/b² = 1
2500/900 - 5625/b² = 1
2500/900 - 1 = 5625/b²
1600/900 = 5625/b²
b² = 5625 × 900/1600 = 3164.06
b ≈ 56.25
Step 3: Find radius at top (y = 75).
x²/900 - 75²/3164.06 = 1
x²/900 - 5625/3164.06 = 1
x²/900 - 1.778 = 1
x²/900 = 2.778
x² = 2500
x = 50 meters
Answer: The radius at the top is 50 meters (same as the base, due to symmetry).
Example 23: Sonic Boom
When a supersonic aircraft breaks the sound barrier, it creates a shock wave in the shape of a cone. The intersection of this cone with the ground forms a hyperbola. If the aircraft is traveling at Mach 2 (twice the speed of sound) at an altitude of 10,000 feet, the resulting hyperbola has the equation x²/10000 - y²/30000 = 1 (in feet). How far from directly below the aircraft will the sonic boom first be heard?
Solution:
Step 1: Identify the relevant distance.
The sonic boom is first heard at the vertices of the hyperbola.
Step 2: Find a.
a² = 10,000
a = 100 feet
Answer: The sonic boom will first be heard 100 feet from directly below the aircraft.
Example 24: Satellite Dish Placement
Two satellite dishes are positioned as the foci of a hyperbolic reflector at (-8, 0) and (8, 0) meters. Signals reflected off the hyperbola satisfy the equation x²/16 - y²/48 = 1. What is the distance between the two vertices of the reflector?
Solution:
Step 1: Identify a from the equation.
a² = 16
a = 4 meters
Step 2: Calculate the distance between vertices.
Vertices are at (±4, 0).
Distance = 4 - (-4) = 8 meters
Answer: The distance between vertices is 8 meters.
Example 25: Particle Physics
In a particle accelerator, the path of a deflected particle follows a hyperbolic trajectory. The equation is given by y²/25 - x²/144 = 1 (in centimeters). At what value of y does the particle cross the line x = 12?
Solution:
Step 1: Substitute x = 12 into the equation.
y²/25 - 12²/144 = 1
y²/25 - 144/144 = 1
y²/25 - 1 = 1
y²/25 = 2
y² = 50
y = ±√50 = ±5√2 ≈ ±7.07
Answer: The particle crosses at y = ±5√2 ≈ ±7.07 cm.
Check Your Understanding
Question 9: In a LORAN system, two stations are 300 km apart. If a ship is 80 km closer to one station, what is the value of a for the hyperbola?
Answer: a = 40 km
Explanation: The difference of distances is 2a = 80 km, so a = 40 km.
Question 10: For the hyperbola x²/36 - y²/64 = 1, what are the coordinates of the foci?
Answer: (±10, 0)
Explanation: a² = 36, b² = 64, so c² = 36 + 64 = 100, giving c = 10. Foci at (±10, 0).
Key Takeaways
- A hyperbola is defined by the absolute value of the difference of distances from two foci being constant.
- Horizontal hyperbolas have the form (x - h)²/a² - (y - k)²/b² = 1; vertical have (y - k)²/a² - (x - h)²/b² = 1.
- The relationship c² = a² + b² is used to find the foci (note: this is different from ellipses!).
- Asymptotes are y - k = ±(b/a)(x - h) for horizontal and y - k = ±(a/b)(x - h) for vertical hyperbolas.
- The central rectangle method uses vertices and co-vertices to graph hyperbolas and asymptotes accurately.
- Eccentricity e = c/a measures the "openness" of a hyperbola; e > 1 for all hyperbolas.
- Complete the square to convert general form equations to standard form.
- Applications include LORAN navigation, hyperbolic mirrors, cooling towers, and sonic booms.
- Remember: Ellipse uses addition and c² = a² - b²; Hyperbola uses subtraction and c² = a² + b².