Ellipse
See how to solve an ellipse
(major/minor axis, foci, equation).
5 examples and their solutions.
Definition
PF + PF' = (constant)
PF + PF' = (constant, major axis).
F, F': foci
Ellipse: x2a2 + y2b2 = 1 (a > b)
Equation
x2a2 + y2b2 = 1 (a > b)
a2 - b2 = c2
Major Axis: 2a
Minor Axis: 2b
Foci: (c, 0), (-c, 0)
The major axis is the longest diameter: 2a.a2 - b2 = c2
Major Axis: 2a
Minor Axis: 2b
Foci: (c, 0), (-c, 0)
The minor axis is the shortest diameter: 2b.
a > b
Then the foci are located horizontally.
By using a2 - b2 = c2,
you can find the foci (±c, 0).
Example
x225 + y216 = 1
1. Major axis?
2. Minor axis?
3. Foci?
Solution 1. Major axis?
2. Minor axis?
3. Foci?
x225 + y216 = 1
x252 + y242 = 1
1. (major axis) = 2⋅5
= 10
2. (minor axis) = 2⋅4
= 8
3. c2 = 52 - 42
= 25 - 16
= 9
c = ±3
Foci: (3, 0), (-3, 0)
x252 + y242 = 1
1. (major axis) = 2⋅5
= 10
2. (minor axis) = 2⋅4
= 8
3. c2 = 52 - 42
= 25 - 16
= 9
c = ±3
Foci: (3, 0), (-3, 0)
Close
Example
Foci: (4, 0), (-4, 0)
Major axis: 10
Equation of the ellipse?
Solution Major axis: 10
Equation of the ellipse?
c = 4
2a = 10
a = 5
52 - b2 = 42
25 - b2 = 16
-b2 = 16 - 25
-b2 = -9
b2 = 9
x252 + y29 = 1 - [2]
x225 + y29 = 1
[1]
Draw the ellipse
and the foci (4, 0), (-4, 0).
Foci are located horizontally.
So the major axis, 10,
is the horizontal diameter.
So 2a = 10.
and the foci (4, 0), (-4, 0).
Foci are located horizontally.
So the major axis, 10,
is the horizontal diameter.
So 2a = 10.
[2]
a = 5
b2 = 9
So the equation of the ellipse is
x2/52 + y2/9 = 1.
b2 = 9
So the equation of the ellipse is
x2/52 + y2/9 = 1.
Close
Example
Foci: (0, 1), (4, 1)
Major axis: 6
Equation of the ellipse?
Solution Major axis: 6
Equation of the ellipse?
2c = 0 + 4
2c = 4
c = 2
(2, 0) → (4, 1) = (2 + 2, 0 + 1)
(x, y) → (x + 2, y + 1) - [2]
2a = 6
a = 3
32 - b2 = 22
9 - b2 = 4
-b2 = 9 - 4
-b2 = -5
b2 = 5
(x - 2)232 + (y - 1)25 = 1 - [3]
(x - 2)29 + (y - 1)25 = 1
[1]
Draw the ellipse
and the foci (0, 1), (4, 1).
Foci are located horizontally.
So the major axis, 6,
is the horizontal diameter.
So 2a = 6.
and the foci (0, 1), (4, 1).
Foci are located horizontally.
So the major axis, 6,
is the horizontal diameter.
So 2a = 6.
[2]
c = 2
So the right focus should be (2, 0).
But the right focus is (4, 1).
Then there's a translation
(2, 0) → (4, 1).
(4, 1) = (2 + 2, 0 + 1)
So the translation is
(x, y) → (x + 2, y + 1).
So the right focus should be (2, 0).
But the right focus is (4, 1).
Then there's a translation
(2, 0) → (4, 1).
(4, 1) = (2 + 2, 0 + 1)
So the translation is
(x, y) → (x + 2, y + 1).
[3]
a = 3
b2 = 5
(x, y) → (x + 2, y + 1)
So the equation of the ellipse is
(x - 2)2/32 + (y - 1)2/5 = 1.
b2 = 5
(x, y) → (x + 2, y + 1)
So the equation of the ellipse is
(x - 2)2/32 + (y - 1)2/5 = 1.
Close
Ellipse: x2a2 + y2b2 = 1 (a < b)
Equation
x2a2 + y2b2 = 1 (a < b)
b2 - a2 = c2
Major Axis: 2b
Minor Axis: 2a
Foci: (0, c), (0, -c)
b2 - a2 = c2
Major Axis: 2b
Minor Axis: 2a
Foci: (0, c), (0, -c)
Example
9x2+ 4y2 = 36
1. Major axis?
2. Minor axis?
3. Foci?
Solution 1. Major axis?
2. Minor axis?
3. Foci?
9x2+ 4y2 = 36
x24 + y29 = 1 - [1]
x222 + y232 = 1
1. (major axis) = 2⋅3
= 6
2. (minor axis) = 2⋅2
= 4
3. c2 = 32 - 22
= 9 - 4
= 5
c = ±√5
Foci: (0, √5), (0, -√5)
x24 + y29 = 1 - [1]
x222 + y232 = 1
1. (major axis) = 2⋅3
= 6
2. (minor axis) = 2⋅2
= 4
3. c2 = 32 - 22
= 9 - 4
= 5
c = ±√5
Foci: (0, √5), (0, -√5)
[1]
÷36 both sides.
Close
Example
Foci: (0, 2), (0, -2)
Major axis: 8
Equation of the ellipse?
Solution Major axis: 8
Equation of the ellipse?
c = 2
2b = 8
b = 4
42 - a2 = 22
16 - a2 = 4
-a2 = 4 - 16
-a2 = -12
a2 = 12
x212 + y242 = 1 - [2]
x212 + y216 = 1
[1]
Draw the ellipse
and the foci (0, 2), (0, -2).
Foci are located vertically.
So the major axis, 8,
is the vertical diameter.
So 2b = 8.
and the foci (0, 2), (0, -2).
Foci are located vertically.
So the major axis, 8,
is the vertical diameter.
So 2b = 8.
[2]
a2 = 12
b = 4
So the equation of the ellipse is
x2/12 + y2/42 = 1.
b = 4
So the equation of the ellipse is
x2/12 + y2/42 = 1.
Close