Lesson 3: Ellipses
Learning Objectives
- Understand the definition of an ellipse using the two-foci distance property
- Identify the standard form equations for horizontal and vertical ellipses
- Distinguish between horizontal and vertical ellipses using a and b values
- Find the center, vertices, co-vertices, and foci of an ellipse
- Use the relationship c² = a² - b² to find the distance to foci
- Identify and interpret the major axis and minor axis
- Graph ellipses centered at the origin and at other points
- Write the equation of an ellipse given geometric properties
- Calculate and interpret eccentricity (e = c/a) as a measure of elongation
- Apply ellipses to real-world contexts: planetary orbits, architecture, and acoustics
Introduction: What is an Ellipse?
Ellipse: An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (called foci) is constant.
If F₁ and F₂ are the two foci and P is any point on the ellipse, then:
d(P, F₁) + d(P, F₂) = constant
An ellipse looks like an elongated circle or an oval. Unlike a circle, which has one center point equidistant from all points on the circle, an ellipse has two focal points. The shape of an ellipse depends on how far apart these foci are.
Visualizing an Ellipse
You can draw an ellipse using two pins (the foci), a loop of string, and a pencil. Place the pins on a board, loop the string around them, pull it taut with the pencil, and trace around while keeping the string tight. The resulting shape is an ellipse because the total length of string (sum of distances to both foci) remains constant.
Key Components of an Ellipse
- Center: The midpoint between the two foci, denoted (h, k)
- Foci (plural of focus): Two fixed points inside the ellipse, denoted F₁ and F₂
- Major Axis: The longest diameter of the ellipse, passing through both foci
- Minor Axis: The shortest diameter of the ellipse, perpendicular to the major axis
- Vertices: The endpoints of the major axis (the farthest points from the center)
- Co-vertices: The endpoints of the minor axis
- a: The distance from the center to a vertex (semi-major axis length)
- b: The distance from the center to a co-vertex (semi-minor axis length)
- c: The distance from the center to a focus
Important: a is always the larger value
By definition, a > b in an ellipse. The value a is associated with the major axis (the longer axis), and b is associated with the minor axis (the shorter axis). This is crucial for determining whether an ellipse is horizontal or vertical.
Section 1: Standard Form Equations of Ellipses
The equation of an ellipse depends on whether its major axis is horizontal or vertical.
Horizontal Ellipse (Major Axis Parallel to x-axis)
Standard Form (Horizontal):
(x - h)²/a² + (y - k)²/b² = 1
where a > b, and center is (h, k)
For a horizontal ellipse:
- The larger denominator is under the (x - h)² term
- The major axis is horizontal (parallel to the x-axis)
- Vertices are at (h ± a, k)
- Co-vertices are at (h, k ± b)
- Foci are at (h ± c, k), where c² = a² - b²
Vertical Ellipse (Major Axis Parallel to y-axis)
Standard Form (Vertical):
(x - h)²/b² + (y - k)²/a² = 1
where a > b, and center is (h, k)
For a vertical ellipse:
- The larger denominator is under the (y - k)² term
- The major axis is vertical (parallel to the y-axis)
- Vertices are at (h, k ± a)
- Co-vertices are at (h ± b, k)
- Foci are at (h, k ± c), where c² = a² - b²
The Relationship Between a, b, and c
For any ellipse, the relationship between the semi-major axis (a), semi-minor axis (b), and focal distance (c) is:
c² = a² - b²
This relationship comes from the geometry of the ellipse. Note that c is always less than a.
Example 1: Identifying Ellipse Orientation
Determine whether each ellipse is horizontal or vertical:
(a) x²/25 + y²/16 = 1
(b) x²/9 + y²/36 = 1
Solution:
(a) x²/25 + y²/16 = 1
Here, a² = 25 (so a = 5) and b² = 16 (so b = 4)
The larger denominator (25) is under x²
This is a horizontal ellipse
(b) x²/9 + y²/36 = 1
Here, a² = 36 (so a = 6) and b² = 9 (so b = 3)
The larger denominator (36) is under y²
This is a vertical ellipse
Example 2: Finding Center and Axis Lengths (Centered at Origin)
For the ellipse x²/49 + y²/25 = 1, find the center, a, b, and determine if it's horizontal or vertical.
Solution:
Step 1: Identify the center.
Since the equation is in the form x²/a² + y²/b² = 1, the center is at the origin (0, 0).
Step 2: Find a and b.
a² = 49, so a = 7
b² = 25, so b = 5
Step 3: Determine orientation.
Since 49 > 25, the larger denominator is under x²
This is a horizontal ellipse
Answer: Center: (0, 0); a = 7; b = 5; horizontal ellipse.
Example 3: Finding All Key Features (Horizontal Ellipse)
For the ellipse x²/36 + y²/20 = 1, find the center, vertices, co-vertices, and foci.
Solution:
Step 1: Identify a² and b².
a² = 36, so a = 6
b² = 20, so b = √20 = 2√5 ≈ 4.47
Step 2: The center is (0, 0).
Step 3: Since 36 > 20, this is a horizontal ellipse.
Step 4: Find the vertices (on the major axis).
Vertices: (±a, 0) = (±6, 0)
Vertices: (6, 0) and (-6, 0)
Step 5: Find the co-vertices (on the minor axis).
Co-vertices: (0, ±b) = (0, ±2√5)
Co-vertices: (0, 2√5) and (0, -2√5) or approximately (0, 4.47) and (0, -4.47)
Step 6: Find c using c² = a² - b².
c² = 36 - 20 = 16
c = 4
Step 7: Find the foci (on the major axis).
Foci: (±c, 0) = (±4, 0)
Foci: (4, 0) and (-4, 0)
Answer:
- Center: (0, 0)
- Vertices: (6, 0) and (-6, 0)
- Co-vertices: (0, 2√5) and (0, -2√5)
- Foci: (4, 0) and (-4, 0)
Example 4: Finding All Key Features (Vertical Ellipse)
For the ellipse x²/16 + y²/64 = 1, find the center, vertices, co-vertices, and foci.
Solution:
Step 1: Identify a² and b².
Since 64 > 16, we have a² = 64 and b² = 16
a = 8, b = 4
Step 2: The center is (0, 0).
Step 3: Since the larger denominator (64) is under y², this is a vertical ellipse.
Step 4: Find the vertices (on the major axis, which is vertical).
Vertices: (0, ±a) = (0, ±8)
Vertices: (0, 8) and (0, -8)
Step 5: Find the co-vertices (on the minor axis, which is horizontal).
Co-vertices: (±b, 0) = (±4, 0)
Co-vertices: (4, 0) and (-4, 0)
Step 6: Find c using c² = a² - b².
c² = 64 - 16 = 48
c = √48 = 4√3 ≈ 6.93
Step 7: Find the foci (on the major axis, which is vertical).
Foci: (0, ±c) = (0, ±4√3)
Foci: (0, 4√3) and (0, -4√3) or approximately (0, 6.93) and (0, -6.93)
Answer:
- Center: (0, 0)
- Vertices: (0, 8) and (0, -8)
- Co-vertices: (4, 0) and (-4, 0)
- Foci: (0, 4√3) and (0, -4√3)
Example 5: Ellipse Not Centered at Origin (Horizontal)
For the ellipse (x - 3)²/25 + (y + 2)²/9 = 1, find the center, vertices, co-vertices, and foci.
Solution:
Step 1: Identify h, k, a², and b².
h = 3, k = -2 (note: y + 2 = y - (-2))
a² = 25, so a = 5
b² = 9, so b = 3
Step 2: The center is (h, k) = (3, -2).
Step 3: Since 25 > 9 and the larger denominator is under (x - h)², this is a horizontal ellipse.
Step 4: Find the vertices.
Vertices: (h ± a, k) = (3 ± 5, -2)
Vertices: (8, -2) and (-2, -2)
Step 5: Find the co-vertices.
Co-vertices: (h, k ± b) = (3, -2 ± 3)
Co-vertices: (3, 1) and (3, -5)
Step 6: Find c.
c² = a² - b² = 25 - 9 = 16
c = 4
Step 7: Find the foci.
Foci: (h ± c, k) = (3 ± 4, -2)
Foci: (7, -2) and (-1, -2)
Answer:
- Center: (3, -2)
- Vertices: (8, -2) and (-2, -2)
- Co-vertices: (3, 1) and (3, -5)
- Foci: (7, -2) and (-1, -2)
Example 6: Ellipse Not Centered at Origin (Vertical)
For the ellipse (x + 1)²/4 + (y - 5)²/16 = 1, find the center, vertices, co-vertices, and foci.
Solution:
Step 1: Identify h, k, a², and b².
h = -1, k = 5
Since 16 > 4, we have a² = 16 and b² = 4
a = 4, b = 2
Step 2: The center is (h, k) = (-1, 5).
Step 3: Since the larger denominator (16) is under (y - k)², this is a vertical ellipse.
Step 4: Find the vertices (vertical).
Vertices: (h, k ± a) = (-1, 5 ± 4)
Vertices: (-1, 9) and (-1, 1)
Step 5: Find the co-vertices (horizontal).
Co-vertices: (h ± b, k) = (-1 ± 2, 5)
Co-vertices: (1, 5) and (-3, 5)
Step 6: Find c.
c² = a² - b² = 16 - 4 = 12
c = √12 = 2√3 ≈ 3.46
Step 7: Find the foci (vertical).
Foci: (h, k ± c) = (-1, 5 ± 2√3)
Foci: (-1, 5 + 2√3) and (-1, 5 - 2√3)
Or approximately: (-1, 8.46) and (-1, 1.54)
Answer:
- Center: (-1, 5)
- Vertices: (-1, 9) and (-1, 1)
- Co-vertices: (1, 5) and (-3, 5)
- Foci: (-1, 5 + 2√3) and (-1, 5 - 2√3)
Section 2: Graphing Ellipses
To graph an ellipse, follow these steps:
- Identify the center (h, k)
- Determine if the ellipse is horizontal or vertical
- Find a and b
- Plot the center
- Plot the vertices (distance a from center along major axis)
- Plot the co-vertices (distance b from center along minor axis)
- Sketch a smooth oval through these four points
- Optionally, plot the foci inside the ellipse
Example 7: Graphing an Ellipse Centered at Origin
Graph the ellipse x²/25 + y²/9 = 1.
Solution:
Step 1: Center is (0, 0).
Step 2: Since 25 > 9 and 25 is under x², this is a horizontal ellipse.
Step 3: a = 5, b = 3
Step 4: Plot the vertices.
Vertices: (±5, 0) → (5, 0) and (-5, 0)
Step 5: Plot the co-vertices.
Co-vertices: (0, ±3) → (0, 3) and (0, -3)
Step 6: Sketch the ellipse through these four points.
Description of graph: The ellipse is wider (horizontal), stretching 5 units left and right from the origin, and 3 units up and down from the origin.
Example 8: Graphing a Vertical Ellipse
Graph the ellipse x²/4 + y²/25 = 1.
Solution:
Step 1: Center is (0, 0).
Step 2: Since 25 > 4 and 25 is under y², this is a vertical ellipse.
Step 3: a = 5, b = 2
Step 4: Plot the vertices (vertical).
Vertices: (0, ±5) → (0, 5) and (0, -5)
Step 5: Plot the co-vertices (horizontal).
Co-vertices: (±2, 0) → (2, 0) and (-2, 0)
Step 6: Sketch the ellipse through these four points.
Description of graph: The ellipse is taller (vertical), stretching 5 units up and down from the origin, and 2 units left and right from the origin.
Example 9: Graphing an Ellipse Not Centered at Origin
Graph the ellipse (x - 2)²/16 + (y + 1)²/4 = 1.
Solution:
Step 1: Center is (2, -1).
Step 2: Since 16 > 4 and 16 is under (x - 2)², this is a horizontal ellipse.
Step 3: a = 4, b = 2
Step 4: Plot the vertices.
Vertices: (2 ± 4, -1) → (6, -1) and (-2, -1)
Step 5: Plot the co-vertices.
Co-vertices: (2, -1 ± 2) → (2, 1) and (2, -3)
Step 6: Sketch the ellipse through these points.
Description of graph: The ellipse is centered at (2, -1), stretches 4 units left and right from the center, and 2 units up and down from the center.
Section 3: Writing Equations of Ellipses
To write the equation of an ellipse, we need to know its center, the lengths of the semi-major and semi-minor axes (a and b), and its orientation.
Example 10: Writing Equation from Vertices and Co-vertices
Write the equation of the ellipse with center at (0, 0), vertices at (±7, 0), and co-vertices at (0, ±3).
Solution:
Step 1: Identify the center: (0, 0), so h = 0 and k = 0.
Step 2: The vertices (±7, 0) are horizontal from the center.
This means the major axis is horizontal.
Distance from center to vertex: a = 7
Step 3: The co-vertices (0, ±3) are vertical from the center.
Distance from center to co-vertex: b = 3
Step 4: Use the horizontal ellipse form.
(x - h)²/a² + (y - k)²/b² = 1
(x - 0)²/49 + (y - 0)²/9 = 1
Answer: x²/49 + y²/9 = 1
Example 11: Writing Equation from Center and Axes Lengths
Write the equation of an ellipse with center (4, -3), a = 6, b = 2, with vertical major axis.
Solution:
Step 1: Center: h = 4, k = -3
Step 2: Since the major axis is vertical, use the vertical form.
Step 3: a = 6, b = 2
a² = 36, b² = 4
Step 4: Write the equation.
(x - h)²/b² + (y - k)²/a² = 1
(x - 4)²/4 + (y + 3)²/36 = 1
Answer: (x - 4)²/4 + (y + 3)²/36 = 1
Example 12: Writing Equation from Foci and Vertices
Write the equation of an ellipse centered at (0, 0) with foci at (±4, 0) and vertices at (±5, 0).
Solution:
Step 1: Center: (0, 0)
Step 2: The foci and vertices are horizontal, so this is a horizontal ellipse.
Step 3: From the vertices: a = 5
Step 4: From the foci: c = 4
Step 5: Find b using c² = a² - b².
16 = 25 - b²
b² = 9
b = 3
Step 6: Write the equation.
x²/25 + y²/9 = 1
Answer: x²/25 + y²/9 = 1
Example 13: Writing Equation from Center, One Vertex, and One Focus
An ellipse has center at (-2, 3), one vertex at (-2, 8), and one focus at (-2, 6). Write its equation.
Solution:
Step 1: Center: h = -2, k = 3
Step 2: The vertex and focus have the same x-coordinate as the center, so the major axis is vertical.
Step 3: Find a (distance from center to vertex).
a = |8 - 3| = 5
Step 4: Find c (distance from center to focus).
c = |6 - 3| = 3
Step 5: Find b using c² = a² - b².
9 = 25 - b²
b² = 16
b = 4
Step 6: Write the equation (vertical ellipse).
(x + 2)²/16 + (y - 3)²/25 = 1
Answer: (x + 2)²/16 + (y - 3)²/25 = 1
Example 14: Converting from General Form to Standard Form
Write the equation 9x² + 4y² = 36 in standard form and identify the key features.
Solution:
Step 1: Divide both sides by 36 to get 1 on the right side.
9x²/36 + 4y²/36 = 36/36
x²/4 + y²/9 = 1
Step 2: Identify a and b.
Since 9 > 4, we have a² = 9 and b² = 4
a = 3, b = 2
Step 3: The larger denominator is under y², so this is a vertical ellipse centered at the origin.
Answer: Standard form: x²/4 + y²/9 = 1
Center: (0, 0); Vertical ellipse; a = 3, b = 2
Example 15: Converting with Completing the Square
Write the equation 4x² + 9y² - 16x + 18y - 11 = 0 in standard form.
Solution:
Step 1: Group x and y terms.
4x² - 16x + 9y² + 18y = 11
Step 2: Factor out the coefficients of x² and y².
4(x² - 4x) + 9(y² + 2y) = 11
Step 3: Complete the square for x and y.
For x: x² - 4x → add (4/2)² = 4 inside parentheses, so add 4(4) = 16 to right
For y: y² + 2y → add (2/2)² = 1 inside parentheses, so add 9(1) = 9 to right
4(x² - 4x + 4) + 9(y² + 2y + 1) = 11 + 16 + 9
4(x - 2)² + 9(y + 1)² = 36
Step 4: Divide by 36.
(x - 2)²/9 + (y + 1)²/4 = 1
Answer: (x - 2)²/9 + (y + 1)²/4 = 1
Center: (2, -1); Horizontal ellipse; a = 3, b = 2
Section 4: Eccentricity
Eccentricity (e): A measure of how elongated an ellipse is. It is defined as:
e = c/a
where c is the distance from the center to a focus, and a is the distance from the center to a vertex.
For an ellipse, 0 < e < 1. The closer e is to 0, the more circular the ellipse. The closer e is to 1, the more elongated (flatter) the ellipse.
- e = 0: Perfect circle (c = 0, meaning the foci coincide at the center)
- e close to 0 (e.g., 0.1): Nearly circular ellipse
- e = 0.5: Moderately elongated ellipse
- e close to 1 (e.g., 0.9): Highly elongated, narrow ellipse
Example 16: Finding Eccentricity
Find the eccentricity of the ellipse x²/25 + y²/9 = 1.
Solution:
Step 1: Identify a and b.
a² = 25, so a = 5
b² = 9, so b = 3
Step 2: Find c using c² = a² - b².
c² = 25 - 9 = 16
c = 4
Step 3: Calculate eccentricity.
e = c/a = 4/5 = 0.8
Answer: The eccentricity is 0.8. This ellipse is fairly elongated.
Example 17: Finding Eccentricity of Nearly Circular Ellipse
Find the eccentricity of the ellipse x²/100 + y²/96 = 1.
Solution:
Step 1: a² = 100, so a = 10; b² = 96, so b = √96 ≈ 9.8
Step 2: Find c.
c² = 100 - 96 = 4
c = 2
Step 3: Calculate eccentricity.
e = c/a = 2/10 = 0.2
Answer: The eccentricity is 0.2. This ellipse is nearly circular.
Example 18: Finding Equation Given Eccentricity
Write the equation of an ellipse centered at the origin with a = 8, eccentricity e = 0.6, and horizontal major axis.
Solution:
Step 1: Use e = c/a to find c.
0.6 = c/8
c = 4.8
Step 2: Use c² = a² - b² to find b.
(4.8)² = 64 - b²
23.04 = 64 - b²
b² = 40.96
b = 6.4
Step 3: Write the equation (horizontal).
x²/64 + y²/40.96 = 1
Answer: x²/64 + y²/40.96 = 1 (or using fractions: x²/64 + y²/(1024/25) = 1)
Section 5: Applications of Ellipses
Ellipses appear in many real-world contexts, from planetary motion to architecture to acoustics.
Application 1: Planetary Orbits (Kepler's First Law)
Johannes Kepler discovered that planets orbit the Sun in elliptical paths, with the Sun at one focus. The eccentricity of a planet's orbit determines how circular or elongated the orbit is.
Example 19: Earth's Orbital Eccentricity
Earth's orbit around the Sun is an ellipse with the Sun at one focus. The semi-major axis is approximately 149.6 million km, and the semi-minor axis is approximately 149.58 million km. Find the eccentricity of Earth's orbit.
Solution:
Step 1: Identify a and b.
a = 149.6 million km
b = 149.58 million km
Step 2: Find c.
c² = a² - b²
c² = (149.6)² - (149.58)²
c² = 22,380.16 - 22,374.1764
c² ≈ 5.98
c ≈ 2.45 million km
Step 3: Calculate eccentricity.
e = c/a ≈ 2.45/149.6 ≈ 0.0164
Answer: Earth's orbital eccentricity is approximately 0.0164, meaning Earth's orbit is nearly circular.
Example 20: Halley's Comet
Halley's Comet has a highly elliptical orbit with eccentricity e = 0.967. If the semi-major axis is approximately 2.7 billion km, find the distance c from the center to the focus (where the Sun is located).
Solution:
Use e = c/a:
0.967 = c/2.7
c = 0.967 × 2.7
c ≈ 2.61 billion km
Answer: The Sun is approximately 2.61 billion km from the center of the orbit. This high eccentricity explains why Halley's Comet has such a long, narrow orbit.
Application 2: Whispering Galleries (Acoustic Property)
Elliptical rooms have a special acoustic property: sound waves originating from one focus reflect off the walls and converge at the other focus. This creates "whispering galleries" where a whisper at one focus can be clearly heard at the other focus, even across a large room.
Example 21: Designing a Whispering Gallery
An architect wants to design an elliptical room where two people standing at the foci can hear each other whisper. The room should be 30 feet long (major axis) and 20 feet wide (minor axis). Where should the two listening stations be placed?
Solution:
Step 1: The major axis length is 2a = 30, so a = 15 feet.
Step 2: The minor axis length is 2b = 20, so b = 10 feet.
Step 3: Find c (distance from center to each focus).
c² = a² - b² = 225 - 100 = 125
c = √125 = 5√5 ≈ 11.18 feet
Step 4: The foci are located along the major axis.
Answer: The two listening stations should be placed approximately 11.18 feet from the center of the room along the long axis, or about 22.36 feet apart from each other.
Application 3: Architecture and Design
Example 22: Elliptical Archway
An elliptical archway is to be constructed with a width of 12 feet at the base and a height of 8 feet at the center. Find the equation of the ellipse if the center is at the origin and the major axis is horizontal.
Solution:
Step 1: The width is 12 feet, so the distance from center to side is a = 6 feet.
Step 2: The height is 8 feet, so the distance from center to top is b = 8 feet.
Wait! Since 8 > 6, we actually have a vertical ellipse with a = 8 and b = 6.
Step 3: Write the equation (vertical ellipse).
x²/36 + y²/64 = 1
Answer: The equation is x²/36 + y²/64 = 1.
Example 23: Elliptical Pool Table
An elliptical pool table has foci located 3 feet from the center along the major axis. If the table is 10 feet long, how wide is it?
Solution:
Step 1: The table is 10 feet long, so 2a = 10, giving a = 5 feet.
Step 2: The foci are 3 feet from the center, so c = 3 feet.
Step 3: Use c² = a² - b² to find b.
9 = 25 - b²
b² = 16
b = 4 feet
Step 4: The width is 2b.
Width = 2(4) = 8 feet
Answer: The table is 8 feet wide. (Interesting property: A ball hit from one focus will bounce off the edge and pass through the other focus!)
Application 4: Engineering and Satellite Dishes
Example 24: Satellite Orbit
A satellite orbits Earth in an elliptical path with Earth at one focus. The closest distance (perigee) is 400 km above Earth's surface, and the farthest distance (apogee) is 2,000 km above Earth's surface. If Earth's radius is 6,371 km, find the semi-major axis of the orbit.
Solution:
Step 1: The perigee distance from Earth's center is 6,371 + 400 = 6,771 km.
Step 2: The apogee distance from Earth's center is 6,371 + 2,000 = 8,371 km.
Step 3: The perigee is a - c from the center, and the apogee is a + c from the center.
a - c = 6,771
a + c = 8,371
Step 4: Add the equations.
2a = 15,142
a = 7,571 km
Answer: The semi-major axis is 7,571 km.
Example 25: Medical Application - Lithotripsy
Elliptical reflectors are used in lithotripsy (breaking up kidney stones). Shock waves generated at one focus reflect off the elliptical surface and converge at the other focus, where the kidney stone is located. If the reflector has a major axis of 16 cm and minor axis of 12 cm, how far apart should the shock wave generator and the kidney stone be positioned?
Solution:
Step 1: Find a and b.
2a = 16, so a = 8 cm
2b = 12, so b = 6 cm
Step 2: Find c.
c² = a² - b² = 64 - 36 = 28
c = √28 = 2√7 ≈ 5.29 cm
Step 3: The distance between the two foci is 2c.
Distance = 2c = 2(2√7) = 4√7 ≈ 10.58 cm
Answer: The shock wave generator and kidney stone should be positioned approximately 10.58 cm apart.
Check Your Understanding
1. For the ellipse x²/81 + y²/49 = 1, identify the center, a, b, and whether it's horizontal or vertical.
Answer: Center: (0, 0); a = 9; b = 7; Horizontal ellipse
Since 81 > 49 and 81 is under x², the major axis is horizontal.
2. Find the foci of the ellipse x²/25 + y²/64 = 1.
Answer: Foci: (0, √39) and (0, -√39), or approximately (0, 6.24) and (0, -6.24)
This is vertical (64 > 25), so a = 8, b = 5
c² = 64 - 25 = 39, so c = √39
Foci are at (0, ±√39)
3. Write the equation of an ellipse with center (0, 0), vertices at (0, ±10), and co-vertices at (±6, 0).
Answer: x²/36 + y²/100 = 1
Vertical ellipse (vertices on y-axis): a = 10, b = 6
4. Find the center and vertices of (x - 4)²/9 + (y + 2)²/25 = 1.
Answer: Center: (4, -2); Vertices: (4, 3) and (4, -7)
Vertical ellipse (25 > 9), a = 5
Vertices at (h, k ± a) = (4, -2 ± 5)
5. Find the eccentricity of the ellipse x²/100 + y²/36 = 1.
Answer: e = 0.8
a = 10, b = 6
c² = 100 - 36 = 64, so c = 8
e = c/a = 8/10 = 0.8
6. An ellipse has center at (0, 0), a = 5, b = 3, and a horizontal major axis. Write its equation.
Answer: x²/25 + y²/9 = 1
Horizontal: larger denominator under x²
7. Find the co-vertices of the ellipse (x + 3)²/16 + (y - 1)²/36 = 1.
Answer: Co-vertices: (1, 1) and (-7, 1)
Vertical ellipse (36 > 16), center (-3, 1), b = 4
Co-vertices at (h ± b, k) = (-3 ± 4, 1)
8. Write 25x² + 9y² = 225 in standard form.
Answer: x²/9 + y²/25 = 1
Divide both sides by 225:
25x²/225 + 9y²/225 = 1
x²/9 + y²/25 = 1
9. An ellipse has foci at (±3, 0) and vertices at (±5, 0). Find b.
Answer: b = 4
a = 5, c = 3
c² = a² - b² → 9 = 25 - b² → b² = 16 → b = 4
10. A planet's orbit has semi-major axis a = 200 million km and eccentricity e = 0.1. How far is the Sun from the center of the orbit?
Answer: 20 million km
e = c/a → 0.1 = c/200 → c = 20 million km
Key Takeaways
- An ellipse is the set of all points where the sum of distances to two foci is constant
- Standard form for horizontal ellipse: (x - h)²/a² + (y - k)²/b² = 1, where a > b
- Standard form for vertical ellipse: (x - h)²/b² + (y - k)²/a² = 1, where a > b
- The larger denominator determines the direction of the major axis
- a is always associated with the major axis (longer), b with the minor axis (shorter)
- The relationship c² = a² - b² connects the focal distance to the axes lengths
- For horizontal ellipses: vertices at (h ± a, k), foci at (h ± c, k)
- For vertical ellipses: vertices at (h, k ± a), foci at (h, k ± c)
- Eccentricity e = c/a measures elongation, where 0 < e < 1 for ellipses
- e close to 0 means nearly circular; e close to 1 means highly elongated
- Applications include planetary orbits (Kepler's laws), whispering galleries, architecture, and medical devices
- The reflective property: waves from one focus reflect to the other focus