Solved Exercises on Ellipses in Algebra 2

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

Solution: Exercises 1 to 3
1 Standard Form Analysis
Exercise 1
For the ellipse (x²/25) + (y²/9) = 1, find the center, vertices, co-vertices, foci, and eccentricity.
Definition:

Ellipse: The set of all points where the sum of distances to two fixed points (foci) is constant

Standard Form: (x²/a²) + (y²/b²) = 1, where a > b for horizontal major axis

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ (horizontal major axis)}\)
Standard Form of Ellipse
Solution Method:
  1. Identify a² and b² from the standard form
  2. Calculate c using c² = a² - b²
  3. Determine center, vertices, co-vertices, and foci
  4. Calculate eccentricity e = c/a
Given
x²/25+y²/9=1
Parameters
a²=25, b²=9
Result
a=5, b=3, c=4
Step 1: Identify parameters from standard form

(x²/25) + (y²/9) = 1

Comparing to (x²/a²) + (y²/b²) = 1:

a² = 25, so a = 5

b² = 9, so b = 3

Step 2: Calculate focal distance

Since a > b, major axis is horizontal

c² = a² - b² = 25 - 9 = 16

c = 4

Step 3: Find key features

Center: (0, 0)

Vertices: (±a, 0) = (±5, 0)

Co-vertices: (0, ±b) = (0, ±3)

Foci: (±c, 0) = (±4, 0)

Step 4: Calculate eccentricity

e = c/a = 4/5 = 0.8

Center: (0,0), Vertices: (±5,0), Co-vertices: (0,±3), Foci: (±4,0), e = 0.8
Final answer:

Center: (0, 0), Vertices: (±5, 0), Co-vertices: (0, ±3), Foci: (±4, 0), Eccentricity: e = 0.8

Applied rules:

• Relationship: c² = a² - b² for ellipses

• Orientation: Major axis along direction of larger denominator

• Eccentricity: e = c/a, where 0 < e < 1

Tip 1: The largest denominator indicates the major axis direction.
Tip 2: Foci are always closer to center than vertices.
2 Vertical Major Axis
Exercise 2
For the ellipse (x²/16) + (y²/36) = 1, find all key features and determine the orientation of the major axis.
Definition:

Vertical Major Axis: When b > a in the standard form, the major axis is vertical

\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \text{ (vertical major axis)}\)
Standard Form (Vertical Major Axis)
Given
x²/16+y²/36=1
Parameters
b²=16, a²=36
Result
a=6, b=4, c=2√5
Step 1: Identify parameters from standard form

(x²/16) + (y²/36) = 1

Comparing to (x²/b²) + (y²/a²) = 1:

b² = 16, so b = 4

a² = 36, so a = 6

Step 2: Determine orientation

Since a² > b² (36 > 16), the major axis is vertical

Step 3: Calculate focal distance

c² = a² - b² = 36 - 16 = 20

c = √20 = 2√5 ≈ 4.47

Step 4: Find key features

Center: (0, 0)

Vertices: (0, ±a) = (0, ±6)

Co-vertices: (±b, 0) = (±4, 0)

Foci: (0, ±c) = (0, ±2√5)

Step 5: Calculate eccentricity

e = c/a = 2√5/6 = √5/3 ≈ 0.745

Center: (0,0), Vertices: (0,±6), Co-vertices: (±4,0), Foci: (0,±2√5), e ≈ 0.745
Final answer:

Center: (0, 0), Vertices: (0, ±6), Co-vertices: (±4, 0), Foci: (0, ±2√5), Eccentricity: e = √5/3

Applied rules:

• Orientation: Major axis along direction of larger denominator

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

• Foci location: Along major axis, inside the ellipse

Tip 1: Always identify which axis is major by comparing denominators.
Tip 2: For vertical major axis, vertices are at (0, ±a) and foci at (0, ±c).
3 Translated Ellipse
Exercise 3
Find the center, vertices, and foci of the ellipse ((x-2)²/9) + ((y+1)²/4) = 1.
Definition:

Translated Ellipse: An ellipse 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 Ellipse
Given
(x-2)²/9+(y+1)²/4=1
Parameters
a²=9, b²=4
Result
Center: (2,-1)
Step 1: Identify parameters from standard form

((x-2)²/9) + ((y+1)²/4) = 1

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

h = 2, k = -1 (center at (2, -1))

a² = 9, so a = 3

b² = 4, so b = 2

Step 2: Determine orientation

Since a > b (3 > 2), major axis is horizontal

Step 3: Calculate focal distance

c² = a² - b² = 9 - 4 = 5

c = √5 ≈ 2.24

Step 4: Find key features

Center: (2, -1)

Vertices: (h±a, k) = (2±3, -1) = (5, -1) and (-1, -1)

Co-vertices: (h, k±b) = (2, -1±2) = (2, 1) and (2, -3)

Foci: (h±c, k) = (2±√5, -1)

Center: (2,-1), Vertices: (5,-1) and (-1,-1), Foci: (2±√5,-1)
Final answer:

Center: (2, -1), Vertices: (5, -1) and (-1, -1), Foci: (2±√5, -1)

Applied rules:

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

• Vertex calculation: Add/subtract a or b from center coordinates

• Focus calculation: Add/subtract c from center coordinates

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

Ellipse: Set of points where sum of distances to foci is constant

Major Axis: Longest diameter of the ellipse

Minor Axis: Shortest diameter of the ellipse

Foci: Two fixed points inside the ellipse

Eccentricity: Measure of how elongated the ellipse is

Problem-Solving Strategy:
  1. Identify orientation: Determine if major 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, co-vertices, foci
  5. Compute eccentricity: Calculate e = c/a
Common Errors: Confusing major/minor axes, sign errors in center coordinates, incorrect parameter identification.
Exam Tips: Memorize standard forms, practice parameter identification, understand eccentricity interpretation.
Solution: Exercises 4 to 5
4 Real-world Application
Exercise 4
The orbit of a planet around a star is elliptical with the star at one focus. If the semi-major axis is 150 million km and the semi-minor axis is 149 million km, find the distance between the planet and star at perihelion (closest approach) and aphelion (farthest distance).
Definition:

Astronomical Application: Planetary orbits follow elliptical paths with star at one focus

Perihelion/Aphelion: Closest and farthest distances from the star

Given
a=150M km, b=149M km
Calculate
c²=a²-b²
Result
Perihelion: 150-c
Step 1: Identify given parameters

Semi-major axis: a = 150 million km

Semi-minor axis: b = 149 million km

Star is at one focus of the ellipse

Step 2: Calculate focal distance

c² = a² - b² = (150)² - (149)²

c² = 22,500 - 22,201 = 299

c = √299 ≈ 17.29 million km

Step 3: Calculate perihelion and aphelion

Perihelion (closest distance): a - c = 150 - 17.29 = 132.71 million km

Aphelion (farthest distance): a + c = 150 + 17.29 = 167.29 million km

Step 4: Verify eccentricity

Eccentricity: e = c/a = 17.29/150 ≈ 0.115

This is a low eccentricity (nearly circular) orbit

Perihelion: 132.71M km, Aphelion: 167.29M km
Final answer:

Perihelion (closest distance): 132.71 million km, Aphelion (farthest distance): 167.29 million km

Applied rules:

• Astronomical model: Star at one focus of elliptical orbit

• Distance calculation: Perihelion = a - c, Aphelion = a + c

• Focal relationship: c² = a² - b²

Tip 1: At perihelion, planet is closest to focus; at aphelion, farthest from focus.
Tip 2: Eccentricity close to 0 means nearly circular orbit; close to 1 means highly elongated.
5 General Form Conversion
Exercise 5
Convert the equation 4x² + 9y² - 16x + 18y - 11 = 0 to standard form and identify all key features.
Definition:

General Form: Ax² + By² + Cx + Dy + E = 0 for ellipses

Completing the Square: Technique to convert to standard form

Original
4x²+9y²-16x+18y-11=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 = 11

Step 2: Factor out coefficients

4(x² - 4x) + 9(y² + 2y) = 11

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: 1² = 1

Add inside parentheses: (y² + 2y + 1)

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

So add 9 to right side: 9(y² + 2y + 1) = 9(y + 1)²

Step 5: Factor and simplify

4(x - 2)² + 9(y + 1)² = 11 + 16 + 9 = 36

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

Step 6: Identify key features

Center: (2, -1)

a² = 9, so a = 3 (major axis horizontal since a > b)

b² = 4, so b = 2

c² = a² - b² = 9 - 4 = 5, so c = √5

Standard form: (x-2)²/9+(y+1)²/4=1, Center: (2,-1), a=3, b=2, c=√5
Final answer:

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

Applied rules:

• Factoring: Factor coefficients before completing the square

• Completing the square: Account for factored coefficients

• 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 to multiply added values by factored coefficients when balancing the equation.
Comprehensive Guide: Ellipses
\(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
Standard Form (Horizontal Major Axis)
\(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)
Standard Form (Vertical Major Axis)
Key definitions:

Ellipse: The set of all points where the sum of distances to two fixed points (foci) is constant

Major Axis: The longest diameter of the ellipse, length 2a

Minor Axis: The shortest diameter of the ellipse, length 2b

Foci: Two fixed points inside the ellipse, distance c from center

Eccentricity: Measure of how elongated the ellipse is, e = c/a where 0 < e < 1

Complete methodology:
  1. Identify orientation: Compare denominators to determine major 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, co-vertices, foci
  5. Compute eccentricity: Calculate e = c/a
  6. Verify relationships: Check that c < a and 0 < e < 1
Tip 1: The largest denominator in standard form indicates the major axis direction.
Tip 2: Remember that c is always smaller than a in an ellipse.
Tip 3: Eccentricity ranges from 0 (circle) to nearly 1 (highly elongated ellipse).
Tip 4: For translated ellipses, add parameters to center coordinates to find vertices and foci.
Common errors: Confusing major and minor axes, sign errors in center coordinates, incorrect focal distance calculation, mixing up horizontal and vertical orientations.
Exam preparation: Memorize standard forms, practice converting between general and standard forms, work on real-world applications.
Essential properties to know:

• Relationship: c² = a² - b² (not c² = a² + b² like hyperbolas)

• Eccentricity: e = c/a, where 0 < e < 1

• Orientation: Major axis along larger denominator's variable

• Sum of distances: For any point P on ellipse, PF₁ + PF₂ = 2a

Visual Understanding: Ellipse Properties
Exercise 6: Ellipse Comparison
Compare different ellipses:
• Circle: x² + y² = 25 (eccentricity = 0)
• Low eccentricity: x²/25 + y²/24 = 1 (eccentricity ≈ 0.2)
• Medium eccentricity: x²/25 + y²/9 = 1 (eccentricity = 0.8)
• High eccentricity: x²/25 + y²/1 = 1 (eccentricity ≈ 0.98)

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

  • Eccentricity near 0: Nearly circular
  • Eccentricity near 1: Highly elongated
  • Circle is an ellipse with e = 0
  • All ellipses have 0 ≤ e < 1

Questions & Answers

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

Answer: Look at the denominators in the standard form:

  • Horizontal major axis: (x²/a²) + (y²/b²) = 1 where a > b
  • Vertical major axis: (x²/b²) + (y²/a²) = 1 where a > b

The major axis is along the variable with the larger denominator. For example, in (x²/25) + (y²/9) = 1, since 25 > 9, the major axis is horizontal. In (x²/9) + (y²/25) = 1, since 25 > 9, the major axis is vertical.

Remember that 'a' is always associated with the major axis, so it's the larger of the two denominators when squared.

Question: What is eccentricity and what does it tell us about an ellipse?

Answer: Eccentricity (e) measures how "stretched out" an ellipse is:

  • Formula: e = c/a, where c is the focal distance and a is the semi-major axis
  • Range: 0 ≤ e < 1 for ellipses
  • Circle: e = 0 (perfectly round)
  • Low e: Nearly circular ellipse
  • High e: Very elongated ellipse

An eccentricity of 0 represents a perfect circle. As eccentricity approaches 1, the ellipse becomes more elongated. Planetary orbits have low eccentricity (close to circular), while some comets have highly eccentric orbits.

The closer e is to 1, the flatter and more stretched the ellipse becomes.

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

Answer: Ellipses have numerous practical applications:

  • Astronomy: Planetary orbits around stars are elliptical
  • Physics: Describing electron orbits in atoms
  • Engineering: Design of bridges, arches, and tunnels
  • Medicine: Lithotripsy (kidney stone treatment) uses elliptical reflectors
  • Acoustics: Whispering galleries use elliptical shapes

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

Understanding ellipses 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 ellipses is:

c² = a² - b²

Where:

  • a: Semi-major axis (longest radius)
  • b: Semi-minor axis (shortest radius)
  • c: Distance from center to each focus

Remember that 'a' is always the largest of the three, and 'c' is always inside the ellipse (c < a). This is different from hyperbolas where c² = a² + b².

A helpful memory device: "c² = a² - b²" (the minus sign reminds you that c is smaller than a).

Also remember that eccentricity e = c/a, which must be between 0 and 1 for ellipses.

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

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 - 11 = 0:
1. Group: 4x² - 16x + 9y² + 18y = 11
2. Factor: 4(x² - 4x) + 9(y² + 2y) = 11
3. Complete: 4(x² - 4x + 4) + 9(y² + 2y + 1) = 11 + 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!