Solved Exercises on Hyperbolas in Algebra 2

Master hyperbolas: equations, foci, vertices, asymptotes, eccentricity, and problem-solving techniques through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Horizontal Transverse Axis
Exercise 1
For the hyperbola (x²/16) - (y²/9) = 1, find the center, vertices, foci, asymptotes, and eccentricity.
Definition:

Hyperbola: The set of all points where the absolute difference of distances to two fixed points (foci) is constant

Standard Form: (x²/a²) - (y²/b²) = 1, where a > 0, b > 0

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ (horizontal transverse axis)}\)
Standard Form of Hyperbola
Solution Method:
  1. Identify a² and b² from the standard form
  2. Calculate c using c² = a² + b²
  3. Determine center, vertices, foci, and asymptotes
  4. Calculate eccentricity e = c/a
Given
x²/16-y²/9=1
Parameters
a²=16, b²=9
Result
a=4, b=3, c=5
Step 1: Identify parameters from standard form

(x²/16) - (y²/9) = 1

Comparing to (x²/a²) - (y²/b²) = 1:

a² = 16, so a = 4

b² = 9, so b = 3

Step 2: Calculate focal distance

c² = a² + b² = 16 + 9 = 25

c = 5

Step 3: Find key features

Center: (0, 0)

Vertices: (±a, 0) = (±4, 0)

Foci: (±c, 0) = (±5, 0)

Step 4: Find asymptotes

For horizontal transverse axis: y = ±(b/a)x

Asymptotes: y = ±(3/4)x

Step 5: Calculate eccentricity

e = c/a = 5/4 = 1.25

Center: (0,0), Vertices: (±4,0), Foci: (±5,0), Asymptotes: y = ±(3/4)x, e = 1.25
Final answer:

Center: (0, 0), Vertices: (±4, 0), Foci: (±5, 0), Asymptotes: y = ±(3/4)x, Eccentricity: e = 1.25

Applied rules:

• Relationship: c² = a² + b² for hyperbolas (different from ellipses)

• Asymptotes: y = ±(b/a)x for horizontal transverse axis

• Eccentricity: e = c/a > 1 for hyperbolas

Tip 1: The positive term indicates the direction of the transverse axis.
Tip 2: For hyperbolas, c is always larger than both a and b.
2 Vertical Transverse Axis
Exercise 2
For the hyperbola (y²/25) - (x²/16) = 1, find all key features and determine the orientation of the transverse axis.
Definition:

Vertical Transverse Axis: When y² term is positive in the standard form

\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \text{ (vertical transverse axis)}\)
Standard Form (Vertical Transverse Axis)
Given
y²/25-x²/16=1
Parameters
a²=25, b²=16
Result
a=5, b=4, c=√41
Step 1: Identify parameters from standard form

(y²/25) - (x²/16) = 1

Comparing to (y²/a²) - (x²/b²) = 1:

a² = 25, so a = 5

b² = 16, so b = 4

Step 2: Determine orientation

Since y² term is positive, the transverse axis is vertical

Step 3: Calculate focal distance

c² = a² + b² = 25 + 16 = 41

c = √41 ≈ 6.4

Step 4: Find key features

Center: (0, 0)

Vertices: (0, ±a) = (0, ±5)

Foci: (0, ±c) = (0, ±√41)

Step 5: Find asymptotes

For vertical transverse axis: y = ±(a/b)x

Asymptotes: y = ±(5/4)x

Step 6: Calculate eccentricity

e = c/a = √41/5 ≈ 1.28

Center: (0,0), Vertices: (0,±5), Foci: (0,±√41), Asymptotes: y = ±(5/4)x, e ≈ 1.28
Final answer:

Center: (0, 0), Vertices: (0, ±5), Foci: (0, ±√41), Asymptotes: y = ±(5/4)x, Eccentricity: e = √41/5

Applied rules:

• Orientation: Transverse axis along positive term's variable

• Vertical transverse axis: Vertices have y-coordinates ±a

• Asymptotes: y = ±(a/b)x for vertical transverse axis

Tip 1: Always identify which term is positive to determine transverse axis direction.
Tip 2: For vertical transverse axis, vertices are at (0, ±a) and foci at (0, ±c).
3 Translated Hyperbola
Exercise 3
Find the center, vertices, foci, and asymptotes of the hyperbola ((x-3)²/9) - ((y+2)²/4) = 1.
Definition:

Translated Hyperbola: A hyperbola with center at (h, k) instead of origin

Standard Form: ((x-h)²/a²) - ((y-k)²/b²) = 1

\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
Standard Form of Translated Hyperbola
Given
(x-3)²/9-(y+2)²/4=1
Parameters
a²=9, b²=4
Result
Center: (3,-2)
Step 1: Identify parameters from standard form

((x-3)²/9) - ((y+2)²/4) = 1

Comparing to ((x-h)²/a²) - ((y-k)²/b²) = 1:

h = 3, k = -2 (center at (3, -2))

a² = 9, so a = 3

b² = 4, so b = 2

Step 2: Calculate focal distance

c² = a² + b² = 9 + 4 = 13

c = √13 ≈ 3.61

Step 3: Find key features

Center: (3, -2)

Since x-term is positive, transverse axis is horizontal

Vertices: (h±a, k) = (3±3, -2) = (6, -2) and (0, -2)

Foci: (h±c, k) = (3±√13, -2)

Step 4: Find asymptotes

For horizontal transverse axis: y - k = ±(b/a)(x - h)

Asymptotes: y - (-2) = ±(2/3)(x - 3)

Asymptotes: y + 2 = ±(2/3)(x - 3)

Asymptotes: y = ±(2/3)(x - 3) - 2

Step 5: Calculate eccentricity

e = c/a = √13/3 ≈ 1.20

Center: (3,-2), Vertices: (6,-2) and (0,-2), Foci: (3±√13,-2), Asymptotes: y = ±(2/3)(x-3)-2
Final answer:

Center: (3, -2), Vertices: (6, -2) and (0, -2), Foci: (3±√13, -2), Asymptotes: y = ±(2/3)(x-3)-2

Applied rules:

• Translation: Center moves from origin to (h, k)

• Vertex calculation: Add/subtract a from center coordinates

• Asymptote equation: Adjust for center translation

Tip 1: The center coordinates are the opposites of the numbers in parentheses.
Tip 2: For translated hyperbolas, add parameters to center coordinates to find vertices and foci.
Key Formulas and Properties
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ (horizontal transverse axis)}\)
Standard Form (Horizontal)
\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \text{ (vertical transverse axis)}\)
Standard Form (Vertical)
\(c^2 = a^2 + b^2 \text{ and } e = \frac{c}{a}\)
Focal Distance and Eccentricity
Horizontal
Transverse Axis
Vertices: (±a,0), Asymptotes: y = ±(b/a)x
Vertical
Transverse Axis
Vertices: (0,±a), Asymptotes: y = ±(a/b)x
Eccentricity
Shape Measure
e > 1, e = 1 for parabola
Key Definitions:

Hyperbola: Set of points where absolute difference of distances to foci is constant

Transverse Axis: Axis that passes through the vertices

Conjugate Axis: Axis perpendicular to transverse axis

Foci: Two fixed points outside the hyperbola

Asymptotes: Lines that the hyperbola approaches but never touches

Eccentricity: Measure of how "open" the hyperbola is

Problem-Solving Strategy:
  1. Identify orientation: Determine if transverse axis is horizontal or vertical
  2. Extract parameters: Identify a, b, h, k from standard form
  3. Calculate c: Use c² = a² + b²
  4. Find key features: Calculate vertices, foci, and asymptotes
  5. Compute eccentricity: Calculate e = c/a
Common Errors: Confusing transverse/conjugate axes, sign errors in center coordinates, incorrect parameter identification, mixing up asymptote formulas.
Exam Tips: Memorize standard forms, practice parameter identification, understand asymptote formulas, know eccentricity properties.
Solution: Exercises 4 to 5
4 Real-world Application
Exercise 4
The cooling tower of a nuclear power plant has a hyperbolic cross-section with the equation (x²/100) - (y²/64) = 1 (measurements in meters). If the tower is 50 meters tall, find the width at the top and bottom of the tower.
Definition:

Hyperbolic Structure: Cooling towers use hyperbolic shape for structural efficiency and stability

Given
x²/100-y²/64=1
Tower height
50m
Result
Width at top: 40m
Step 1: Identify parameters from the equation

(x²/100) - (y²/64) = 1

a² = 100, so a = 10

b² = 64, so b = 8

Step 2: Determine the orientation

Since x² term is positive, transverse axis is horizontal

This means the hyperbola opens left and right

Step 3: Set up the coordinate system

Assume the center of the hyperbola is at ground level (y = 0)

Top of tower: y = 50

Bottom of tower: y = 0

Step 4: Find width at bottom (y = 0)

(x²/100) - (0²/64) = 1

x²/100 = 1

x² = 100

x = ±10

Width at bottom = 2|10| = 20 meters

Step 5: Find width at top (y = 50)

(x²/100) - (50²/64) = 1

(x²/100) - (2500/64) = 1

(x²/100) = 1 + 2500/64

(x²/100) = (64 + 2500)/64 = 2564/64

x² = 100 × (2564/64) = 256400/64 = 4006.25

x = ±√4006.25 ≈ ±63.3

Width at top = 2|63.3| = 126.6 meters

Width at bottom: 20m, Width at top: 126.6m
Final answer:

Width at bottom: 20 meters, Width at top: 126.6 meters

Applied rules:

• Real-world modeling: Translate physical situation to mathematical equation

• Substitution: Plug y-values into equation to find x-values

• Width calculation: Distance between two x-values at same y-value

Tip 1: Set up a coordinate system that makes the problem easier to solve.
Tip 2: Always consider the practical meaning of your solutions.
5 General Form Conversion
Exercise 5
Convert the equation 4x² - 9y² - 16x + 18y - 29 = 0 to standard form and identify all key features.
Definition:

General Form: Ax² + By² + Cx + Dy + E = 0 where A and B have opposite signs

Completing the Square: Technique to convert to standard form

Original
4x²-9y²-16x+18y-29=0
Complete squares
4(x-2)²-9(y-1)²=36
Standard form
(x-2)²/9-(y-1)²/4=1
Step 1: Group x and y terms

4x² - 16x - 9y² + 18y = 29

(4x² - 16x) - (9y² - 18y) = 29

Step 2: Factor out coefficients

4(x² - 4x) - 9(y² - 2y) = 29

Step 3: Complete the square for x

Take half of coefficient: -4/2 = -2

Square it: (-2)² = 4

Add inside parentheses: (x² - 4x + 4)

But since it's multiplied by 4: 4×4 = 16 added to left side

So add 16 to right side: 4(x² - 4x + 4) = 4(x - 2)²

Step 4: Complete the square for y

Take half of coefficient: -2/2 = -1

Square it: (-1)² = 1

Add inside parentheses: (y² - 2y + 1)

But since it's multiplied by -9: -9×1 = -9 added to left side

So subtract 9 from right side: -9(y² - 2y + 1) = -9(y - 1)²

Step 5: Factor and simplify

4(x - 2)² - 9(y - 1)² = 29 + 16 - (-9) = 29 + 16 + 9 = 54

Wait, let's recalculate:

4(x² - 4x) - 9(y² - 2y) = 29

4(x² - 4x + 4) - 9(y² - 2y + 1) = 29 + 4(4) - 9(1) = 29 + 16 - 9 = 36

4(x - 2)² - 9(y - 1)² = 36

Step 6: Divide by constant to get standard form

Divide by 36: (x - 2)²/9 - (y - 1)²/4 = 1

Step 7: Identify key features

Center: (2, 1)

a² = 9, so a = 3 (transverse axis horizontal since x-term is positive)

b² = 4, so b = 2

c² = a² + b² = 9 + 4 = 13, so c = √13

Standard form: (x-2)²/9-(y-1)²/4=1, Center: (2,1), a=3, b=2, c=√13
Final answer:

Standard form: (x - 2)²/9 - (y - 1)²/4 = 1, Center: (2, 1), Vertices: (5, 1) and (-1, 1), Foci: (2±√13, 1)

Applied rules:

• Factoring: Factor coefficients before completing the square

• Completing the square: Account for signs when balancing equation

• Standard form: Divide by constant to get 1 on right side

Tip 1: Always factor coefficients of squared terms before completing the square.
Tip 2: Remember that negative coefficients affect the balance when completing the square.
Comprehensive Guide: Hyperbolas
\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
Standard Form (Horizontal Transverse Axis)
\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)
Standard Form (Vertical Transverse Axis)
Key definitions:

Hyperbola: The set of all points where the absolute difference of distances to two fixed points (foci) is constant

Transverse Axis: The axis that passes through the vertices, length 2a

Conjugate Axis: The axis perpendicular to the transverse axis, length 2b

Foci: Two fixed points outside the hyperbola, distance c from center

Asymptotes: Lines that the hyperbola approaches but never intersects

Eccentricity: Measure of how "open" the hyperbola is, e = c/a where e > 1

Complete methodology:
  1. Identify orientation: Compare signs of x² and y² terms to determine transverse axis direction
  2. Extract parameters: Identify a, b, h, k from standard form
  3. Calculate focal distance: Use c² = a² + b²
  4. Find key features: Calculate center, vertices, foci, and asymptotes
  5. Compute eccentricity: Calculate e = c/a
  6. Verify relationships: Check that c > a and e > 1
Tip 1: The positive term indicates the direction of the transverse axis.
Tip 2: Remember that c is always larger than both a and b in hyperbolas.
Tip 3: Eccentricity is always greater than 1 for hyperbolas.
Tip 4: For translated hyperbolas, add parameters to center coordinates to find vertices and foci.
Common errors: Confusing transverse and conjugate axes, sign errors in center coordinates, incorrect focal distance calculation, mixing up asymptote formulas for horizontal and vertical transverse axes.
Exam preparation: Memorize standard forms, practice converting between general and standard forms, work on real-world applications, master asymptote equations.
Essential properties to know:

• Relationship: c² = a² + b² (different from ellipses: c² = a² - b²)

• Eccentricity: e = c/a, where e > 1

• Orientation: Transverse axis along positive term's variable

• Asymptotes: y = ±(b/a)x (horizontal) or y = ±(a/b)x (vertical)

Visual Understanding: Hyperbola Properties
Exercise 6: Hyperbola Comparison
Compare different hyperbolas:
• Low eccentricity: x²/4 - y²/1 = 1 (e ≈ 1.12)
• Medium eccentricity: x²/4 - y²/4 = 1 (e ≈ 1.41)
• High eccentricity: x²/4 - y²/16 = 1 (e ≈ 2.24)
• Asymptotes approach axes: x²/4 - y²/64 = 1 (e ≈ 4.03)

Analysis: The visualization shows how eccentricity affects the shape of hyperbolas.

  • Low eccentricity: Hyperbola opens more slowly
  • High eccentricity: Hyperbola opens more rapidly
  • Asymptotes determine the limiting behavior
  • All hyperbolas have e > 1

Questions & Answers

Question: How do I know if the transverse axis is horizontal or vertical?

Answer: Look at the signs in the standard form:

  • Horizontal transverse axis: (x²/a²) - (y²/b²) = 1 (positive x² term)
  • Vertical transverse axis: (y²/a²) - (x²/b²) = 1 (positive y² term)

The transverse axis is along the variable with the positive term. For example, in (x²/25) - (y²/9) = 1, since x² is positive, the transverse axis is horizontal. In (y²/25) - (x²/9) = 1, since y² is positive, the transverse axis is vertical.

The transverse axis always passes through the vertices of the hyperbola.

Question: What are asymptotes and how do I find them?

Answer: Asymptotes are lines that the hyperbola approaches but never touches:

  • Horizontal transverse axis: y - k = ±(b/a)(x - h)
  • Vertical transverse axis: y - k = ±(a/b)(x - h)

For a hyperbola centered at (h, k), the asymptotes pass through the center and have slopes determined by the ratio of the semi-axes.

For example, for (x²/9) - (y²/4) = 1, the asymptotes are y = ±(2/3)x. For (y²/9) - (x²/4) = 1, the asymptotes are y = ±(3/2)x.

Asymptotes help define the limiting behavior of the hyperbola as x or y approaches infinity.

Question: What are some real-world applications of hyperbolas? Why is it important to learn this?

Answer: Hyperbolas have numerous practical applications:

  • Architecture: Cooling towers, bridges, and architectural structures
  • Physics: Gravitational slingshot trajectories for spacecraft
  • Navigation: LORAN (Long Range Navigation) systems
  • Optics: Hyperbolic mirrors and lenses
  • Acoustics: Sound reflection patterns

Learning hyperbolas develops understanding of curved geometric shapes and their properties. This knowledge is fundamental to physics, engineering, astronomy, and architecture. The mathematical techniques used in hyperbolas (parameter identification, equation manipulation) are essential for higher mathematics.

Understanding hyperbolas helps in modeling real-world phenomena and solving practical problems in science and engineering.

Question: I sometimes confuse the relationship between a, b, and c. What's the correct formula?

Answer: The relationship for hyperbolas is:

c² = a² + b²

Where:

  • a: Semi-transverse axis (half the distance between vertices)
  • b: Semi-conjugate axis
  • c: Distance from center to each focus

Remember that 'c' is always the largest of the three in hyperbolas (c > a and c > b). This is different from ellipses where c² = a² - b².

A helpful memory device: "c² = a² + b²" (the plus sign reminds you that c is larger than both a and b).

Also remember that eccentricity e = c/a, which is always greater than 1 for hyperbolas.

Question: How do I convert from general form to standard form for hyperbolas?

Answer: Here's the systematic approach:

  1. Group terms: Put x terms together and y terms together
  2. Factor coefficients: Factor out coefficients of x² and y² terms
  3. Complete the square: For each group, take half the linear coefficient, square it, and add inside parentheses
  4. Balance equation: Whatever you add inside parentheses, account for it outside (multiply by factored coefficient)
  5. Divide by constant: Divide entire equation by the constant on the right side to get 1

For example, with 4x² - 9y² - 16x + 18y - 29 = 0:
1. Group: 4x² - 16x - 9y² + 18y = 29
2. Factor: 4(x² - 4x) - 9(y² - 2y) = 29
3. Complete: 4(x² - 4x + 4) - 9(y² - 2y + 1) = 29 + 4(4) - 9(1)
4. Factor: 4(x - 2)² - 9(y - 1)² = 36
5. Divide: (x - 2)²/9 - (y - 1)²/4 = 1

Always remember to account for factored coefficients when balancing the equation!