Quantity	Meaning
a	Semi-major axis length
b	Semi-minor axis length
c	Distance from the center to either focus
2a	Major axis length; also the constant focal-distance sum
2b	Minor axis length
2c	Distance between the two foci

3. The relationship between a, b, and c

For an ellipse:

\[ a^2=b^2+c^2. \]

Equivalently:

\[ c^2=a^2-b^2. \]

Therefore:

\[ c=\sqrt{a^2-b^2}. \]

Why does this relationship hold?

Consider a horizontal ellipse centered at the origin.

Its foci are:

\[ F_1=(-c,0),\qquad F_2=(c,0). \]

The upper co-vertex is:

\[ P=(0,b). \]

By symmetry, P is equally far from both foci.

Since the two distances add to 2a, each distance equals a.

The distance formula gives:

\[ PF_2=\sqrt{c^2+b^2}=a. \]

Squaring:

\[ c^2+b^2=a^2. \]

Remember: The relationship uses squares. In general, a is not equal to b + c.

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4. Standard equations centered at the origin

4.1 Horizontal major axis

\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b>0. \]

The larger denominator is under x², so the ellipse extends farther horizontally.

Its properties are:

• Center: (0, 0).
• Vertices: (−a, 0) and (a, 0).
• Co-vertices: (0, −b) and (0, b).
• Foci: (−c, 0) and (c, 0).

4.2 Vertical major axis

\[ \frac{x^2}{b^2}+\frac{y^2}{a^2}=1, \qquad a>b>0. \]

The larger denominator is under y², so the ellipse extends farther vertically.

Its properties are:

• Center: (0, 0).
• Vertices: (0, −a) and (0, a).
• Co-vertices: (−b, 0) and (b, 0).
• Foci: (0, −c) and (0, c).

Comparison table

Property	Horizontal major axis	Vertical major axis
Standard equation	x²/a² + y²/b² = 1	x²/b² + y²/a² = 1
Larger denominator	Under x²	Under y²
Vertices	(−a, 0), (a, 0)	(0, −a), (0, a)
Co-vertices	(0, −b), (0, b)	(−b, 0), (b, 0)
Foci	(−c, 0), (c, 0)	(0, −c), (0, c)

Important: The denominator under x² is not automatically a². By convention, a² is always the larger denominator.

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5. Eccentricity

Eccentricity measures the elongation of an ellipse.

It is defined by:

\[ e=\frac ca. \]

For a noncircular ellipse:

\[ 0<e<1. \]

Using the relationship between a, b, and c:

\[ e=\sqrt{1-\frac{b^2}{a^2}}. \]

Interpreting eccentricity

• An eccentricity close to 0 means the ellipse is close to circular.
• An eccentricity close to 1 means it is more elongated.

If a = b, then c = 0 and e = 0. The equation represents a circle, the special circular case.

Scale does not determine eccentricity

Two ellipses can have different sizes but the same eccentricity.

For example:

• a = 5, b = 3 gives c = 4 and e = 4/5.
• a = 10, b = 6 gives c = 8 and e = 4/5.

The second ellipse is larger, but both have the same axis ratio.

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6. How to sketch an ellipse

For an ellipse in standard form:

1. Identify the center.
2. Find a and b by taking square roots of the denominators.
3. Identify the major-axis direction.
4. Plot the two vertices.
5. Plot the two co-vertices.
6. Draw a smooth closed curve through these four points.
7. Plot the foci inside the curve if required.

For a centered ellipse, the graph is symmetric about both coordinate axes and the origin.

Coordinate limits

For a horizontal ellipse centered at the origin:

x-values: [−a, a].
y-values: [−b, b].

For a vertical ellipse centered at the origin:

x-values: [−b, b].
y-values: [−a, a].

These describe coordinate projections. The entire ellipse is not the graph of a single function y = f(x), because most permitted x-values correspond to two y-values.

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7. Worked example 1: Reading a horizontal ellipse

Find the axes, vertices, foci, and eccentricity of:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1. \]

Step 1: Identify a and b

The larger denominator is 25:

\[ a^2=25,\qquad b^2=9. \]

Therefore:

\[ a=5,\qquad b=3. \]

The major axis is horizontal.

Step 2: Find c

\[ c^2=25-9=16. \]\[ c=4. \]

Step 3: State the properties

• Center: (0, 0).
• Major axis length: 10.
• Minor axis length: 6.
• Vertices: (−5, 0), (5, 0).
• Co-vertices: (0, −3), (0, 3).
• Foci: (−4, 0), (4, 0).

Step 4: Calculate eccentricity

\[ e=\frac45. \]

Verification: The foci lie inside the vertices because 4 < 5.

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8. Worked example 2: Normalize before reading

Find the foci and eccentricity of:

\[ 9x^2+4y^2=36. \]

Step 1: Make the right-hand side equal to 1

Divide every term by 36:

\[ \frac{x^2}{4}+\frac{y^2}{9}=1. \]

Step 2: Identify the major axis

The larger denominator is under y².

Therefore:

\[ a=3,\qquad b=2. \]

The major axis is vertical.

Step 3: Find c

\[ c^2=9-4=5. \]\[ c=\sqrt5. \]

Foci: (0, −√5) and (0, √5).

Step 4: Find eccentricity

\[ e=\frac{\sqrt5}{3}. \]

The vertices are (0, −3) and (0, 3).

Key lesson: Compare the denominators after normalization, not the original coefficients.

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9. Worked example 3: Constructing an equation

An ellipse has foci at (−4, 0) and (4, 0). Its major axis length is 10.

Find its equation.

Step 1: Identify the center and direction

The center is the midpoint of the foci:

Center: (0, 0).

The major axis is horizontal.

Step 2: Identify a and c

The major axis length is 2a:

\[ 2a=10, \]

so:

\[ a=5. \]

The center-to-focus distance is:

\[ c=4. \]

Step 3: Find b²

\[ b^2=a^2-c^2=25-16=9. \]

Step 4: Write the equation

\[ \frac{x^2}{25}+\frac{y^2}{9}=1. \]

Check: The focal separation is 8, less than the distance sum 10, so the ellipse is nondegenerate.

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10. Worked example 4: Using the focal-distance definition

For the ellipse:

\[ \frac{x^2}{36}+\frac{y^2}{20}=1, \]

a point P on the ellipse is 5 units from one focus.

Find its distance from the other focus.

Step 1: Identify a

\[ a^2=36,\qquad a=6. \]

Step 2: Use the constant distance sum

\[ PF_1+PF_2=2a=12. \]

If one distance is 5:

\[ 5+PF_2=12. \]

Distance from the other focus: 7 units.

Key lesson: The constant sum is 2a, not a or 2c.

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11. Worked example 5: Finding an equation from eccentricity

An ellipse is centered at the origin. Its major axis is vertical, its major axis length is 12, and its eccentricity is 2/3.

Find its equation.

Step 1: Find a

\[ 2a=12, \]

so:

\[ a=6. \]

Step 2: Find c

\[ e=\frac ca. \]

Therefore:

\[ c=ae=6\left(\frac23\right)=4. \]

Step 3: Find b²

\[ b^2=a^2-c^2=36-16=20. \]

Step 4: Use the vertical standard form

\[ \frac{x^2}{20}+\frac{y^2}{36}=1. \]

Verification: The larger denominator is under y², as required.

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12. Translated ellipses

An ellipse does not have to be centered at the origin.

For center (h, k), replace x by x − h and y by y − k.

Horizontal major axis

\[ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1. \]

• Vertices: (h − a, k), (h + a, k).
• Co-vertices: (h, k − b), (h, k + b).
• Foci: (h − c, k), (h + c, k).

Vertical major axis

\[ \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1. \]

• Vertices: (h, k − a), (h, k + a).
• Co-vertices: (h − b, k), (h + b, k).
• Foci: (h, k − c), (h, k + c).

Translation changes the position, but not the axis lengths or eccentricity.

Worked example 6: Completing the square

Identify the center, vertices, foci, and eccentricity of:

\[ 9x^2+25y^2-18x+100y-116=0. \]

Group the terms:

\[ 9(x^2-2x)+25(y^2+4y)=116. \]

Complete the squares:

\[ 9\big((x-1)^2-1\big) +25\big((y+2)^2-4\big)=116. \]

Therefore:

\[ 9(x-1)^2+25(y+2)^2=225. \]

Divide by 225:

\[ \frac{(x-1)^2}{25}+\frac{(y+2)^2}{9}=1. \]

Read the center:

Center: (1, −2).

Also:

\[ a=5,\qquad b=3,\qquad c=4. \]

Therefore:

• Vertices: (−4, −2), (6, −2).
• Co-vertices: (1, −5), (1, 1).
• Foci: (−3, −2), (5, −2).
• Eccentricity: 4/5.

Sign check: The expression (y + 2)² means the center’s y-coordinate is −2.

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13. Checking points and simple intersections

For:

\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \]

substitute the coordinates of a point into the left-hand side.

• Result less than 1: inside the ellipse.
• Result equal to 1: on the ellipse.
• Result greater than 1: outside the ellipse.

Worked example 7: Checking a point

Is P = (3, 12/5) on:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1? \]

Substitute:

\[ \frac{3^2}{25}+\frac{(12/5)^2}{9} =\frac9{25}+\frac{16}{25} =1. \]

Therefore, P lies on the ellipse.

Worked example 8: Intersecting a vertical line

Find where x = 3 intersects:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1. \]

Substitute:

\[ \frac9{25}+\frac{y^2}{9}=1. \]\[ \frac{y^2}{9}=\frac{16}{25}. \]\[ y^2=\frac{144}{25}. \]

Therefore:

\[ y=\pm\frac{12}{5}. \]

Intersection points: (3, 12/5) and (3, −12/5).

Remember both signs when solving a squared equation.

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14. Common mistakes and verification

Mistake 1: Treating the denominator as a length

In an equation with denominator 25, the corresponding semi-axis length is 5, not 25.

Mistake 2: Assuming a² is always under x²

The larger denominator is a², regardless of its position.

Mistake 3: Using the wrong focal relationship

For an ellipse:

\[ c^2=a^2-b^2. \]

Do not add a² and b².

Mistake 4: Confusing a with 2a

The semi-major axis is a. The full major axis is 2a.

Mistake 5: Putting the foci on the minor axis

The foci always lie on the major axis.

Mistake 6: Reversing eccentricity

Use:

\[ e=\frac ca, \]

not a/c. A noncircular ellipse must have eccentricity strictly between 0 and 1.

Mistake 7: Reading a translated center with the wrong signs

The center of an equation containing (x + 3)² and (y − 2)² is (−3, 2).

Mistake 8: Calling every same-sign quadratic an ellipse

For example:

\[ x^2+4y^2=-1 \]

has no real points.

Normalize and check the resulting equation before classifying it.

Quick verification checklist

• Is the right-hand side 1?
• Are both denominators positive?
• Is a the larger semi-axis?
• Does c² = a² − b²?
• Are the foci inside the major-axis endpoints?
• Is 0 < e < 1 for a noncircular ellipse?
• Do the vertices satisfy the equation?

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15. Independent practice

Complete these before reading the solutions.

Practice 1

For:

\[ \frac{x^2}{49}+\frac{y^2}{24}=1, \]

find the center, axis lengths, vertices, co-vertices, foci, and eccentricity.

Practice 2

For:

\[ \frac{x^2}{16}+\frac{y^2}{25}=1, \]

find the major-axis direction, foci, and eccentricity.

Practice 3

An ellipse has vertices at (−6, 0) and (6, 0), and foci at (−2, 0) and (2, 0).

Find its equation.

Practice 4

An ellipse centered at the origin has a vertical major axis, semi-major axis length 10, and eccentricity 3/5.

Find its equation.

Practice 5

For an ellipse, the sum of the distances from any point on the curve to its foci is 18.

If one focal distance is 7, find the other.

Practice 6

Convert to standard form and find the foci:

\[ 4x^2+9y^2=36. \]

Practice 7

Find the center and foci of:

\[ \frac{(x+2)^2}{9}+\frac{(y-1)^2}{25}=1. \]

Practice 8

Find all intersection points of y = 2 with:

\[ \frac{x^2}{20}+\frac{y^2}{5}=1. \]

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16. Practice solutions

Solution 1

\[ a=7,\qquad b=\sqrt{24}=2\sqrt6. \]\[ c=\sqrt{49-24}=5. \]

• Center: (0, 0).
• Major axis length: 14.
• Minor axis length: 4√6.
• Vertices: (−7, 0), (7, 0).
• Co-vertices: (0, −2√6), (0, 2√6).
• Foci: (−5, 0), (5, 0).
• Eccentricity: 5/7.

Solution 2

The larger denominator is under y², so the major axis is vertical.

\[ a=5,\qquad b=4,\qquad c=\sqrt{25-16}=3. \]

Foci: (0, −3), (0, 3).

Eccentricity: 3/5.

Solution 3

The center is the origin and the major axis is horizontal.

\[ a=6,\qquad c=2. \]\[ b^2=36-4=32. \]

Therefore:

\[ \frac{x^2}{36}+\frac{y^2}{32}=1. \]

Solution 4

\[ a=10,\qquad c=ae=10\left(\frac35\right)=6. \]\[ b^2=100-36=64. \]

The major axis is vertical:

\[ \frac{x^2}{64}+\frac{y^2}{100}=1. \]

Solution 5

The two focal distances add to 18:

\[ 7+d=18. \]

Other focal distance: 11 units.

Solution 6

Divide by 36:

\[ \frac{x^2}{9}+\frac{y^2}{4}=1. \]\[ a=3,\qquad b=2,\qquad c=\sqrt5. \]

Foci: (−√5, 0), (√5, 0).

Solution 7

Center: (−2, 1).

The major axis is vertical.

\[ a=5,\qquad b=3,\qquad c=4. \]

Move 4 units up and down from the center.

Foci: (−2, −3), (−2, 5).

Solution 8

Substitute y = 2:

\[ \frac{x^2}{20}+\frac45=1. \]\[ \frac{x^2}{20}=\frac15. \]\[ x^2=4. \]

Intersection points: (−2, 2), (2, 2).

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17. Lesson assessment — 03.4 Ellipses Quiz

Suggested time: 12–15 minutes
Questions: 10 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 8/10

This is a lesson-level assessment, not a full examination simulation.

Question 1

For:

\[ \frac{x^2}{64}+\frac{y^2}{36}=1, \]

what is the major axis length?

A. 8
B. 12
C. 16
D. 64

Question 2

Which equation represents an ellipse with a vertical major axis?

A. \(x^2/25+y^2/9=1\)

B. \(x^2/9+y^2/25=1\)

C. \(x^2/16+y^2/16=1\)

D. \(x^2/9-y^2/25=1\)

Question 3

Find the foci of:

\[ \frac{x^2}{25}+\frac{y^2}{16}=1. \]

A. (−3, 0), (3, 0)
B. (0, −3), (0, 3)
C. (−5, 0), (5, 0)
D. (−4, 0), (4, 0)

Question 4

An ellipse has a = 13 and b = 5.

What is its eccentricity?

A. 5/13
B. 12/5
C. 13/12
D. 12/13

Question 5

An ellipse has foci (−3, 0), (3, 0) and major axis length 10.

Which is its equation?

A. \(x^2/10+y^2/7=1\)

B. \(x^2/25+y^2/9=1\)

C. \(x^2/25+y^2/16=1\)

D. \(x^2/16+y^2/25=1\)

Question 6

For:

\[ \frac{x^2}{49}+\frac{y^2}{9}=1, \]

a point on the ellipse is 6 units from one focus.

How far is it from the other focus?

A. 1
B. 8
C. 7
D. 14

Question 7

Find the center of:

\[ \frac{(x-3)^2}{16}+\frac{(y+2)^2}{9}=1. \]

A. (3, −2)
B. (−3, 2)
C. (3, 2)
D. (−3, −2)

Question 8

Find the foci of:

\[ 4x^2+9y^2=36. \]

A. (0, −√5), (0, √5)
B. (−3, 0), (3, 0)
C. (−√13, 0), (√13, 0)
D. (−√5, 0), (√5, 0)

Question 9

An ellipse centered at the origin has a horizontal major axis, a = 5, and eccentricity 3/5.

Which is its equation?

A. \(x^2/25+y^2/9=1\)

B. \(x^2/16+y^2/25=1\)

C. \(x^2/25+y^2/16=1\)

D. \(x^2/5+y^2/4=1\)

Question 10

Find the intersection points of x = 4 with:

\[ \frac{x^2}{25}+\frac{y^2}{9}=1. \]

A. (4, 3), (4, −3)
B. (4, 9/5), (4, −9/5)
C. (4, 12/5), (4, −12/5)
D. There are no real intersection points.

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18. Instructor answer key and assessment solutions

Tutor LMS placement: Put the assessment in a separate Quiz item. Keep this answer key and the explanations out of the student-facing lesson until the attempt is submitted.

Question	Answer	Skill assessed
1	C	Axis lengths
2	B	Major-axis direction
3	A	Foci
4	D	Eccentricity
5	C	Constructing an equation
6	B	Focal-distance definition
7	A	Translated center
8	D	Normalization and foci
9	C	Equation from eccentricity
10	B	Simple line intersections

Answer 1: C

\[ a=\sqrt{64}=8. \]

The major axis length is 2a.

Answer: 16.

Answer 2: B

The larger denominator is 25 and lies under y².

Therefore, the major axis is vertical.

Option C is a circle, with no unique major-axis direction.

Answer 3: A

\[ c^2=25-16=9. \]\[ c=3. \]

The major axis is horizontal.

Foci: (−3, 0), (3, 0).

Answer 4: D

\[ c=\sqrt{13^2-5^2} =\sqrt{169-25}=12. \]\[ e=\frac{12}{13}. \]

Answer 5: C

\[ a=5,\qquad c=3. \]\[ b^2=25-9=16. \]

The horizontal standard equation is:

\[ \frac{x^2}{25}+\frac{y^2}{16}=1. \]

Answer 6: B

\[ 2a=2\sqrt{49}=14. \]

The two distances add to 14.

Other distance: 14 − 6 = 8 units.

Answer 7: A

Compare the squared terms with x − h and y − k.

Center: (3, −2).

Answer 8: D

Normalize:

\[ \frac{x^2}{9}+\frac{y^2}{4}=1. \]\[ c^2=9-4=5. \]

Foci: (−√5, 0), (√5, 0).

Answer 9: C

\[ c=ae=5\left(\frac35\right)=3. \]\[ b^2=25-9=16. \]

Therefore:

\[ \frac{x^2}{25}+\frac{y^2}{16}=1. \]

Answer 10: B

Substitute x = 4:

\[ \frac{16}{25}+\frac{y^2}{9}=1. \]\[ \frac{y^2}{9}=\frac9{25}. \]\[ y^2=\frac{81}{25}. \]

Intersection points: (4, 9/5), (4, −9/5).

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19. Revision guidance

• Missed Questions 1–2: Review semi-axis lengths and major-axis direction.
• Missed Questions 3 or 8: Review normalization and c² = a² − b².
• Missed Questions 4 or 9: Review eccentricity and constructing equations.
• Missed Question 5: Distinguish major axis length from semi-major axis length.
• Missed Question 6: Review the constant focal-distance sum.
• Missed Question 7: Review translated centers and signs.
• Missed Question 10: Review substitution and both square-root signs.

For every incorrect or guessed answer:

1. Identify the mistake.
2. Write a corrected solution.
3. Check the result using a second property or substitution.

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20. Lesson review

You should now be able to:

• Explain an ellipse using its two foci.
• Distinguish a, b, c, 2a, 2b, and 2c.
• Recognize horizontal and vertical standard equations.
• Identify centers, vertices, co-vertices, and foci.
• Use the relationship between the semi-axes and focal distance.
• Calculate and interpret eccentricity.
• Construct an equation from axes, foci, or eccentricity.
• Read translated equations and complete the square.
• Check whether a point lies on an ellipse.
• Solve simple horizontal or vertical line-intersection problems.

The central relationships are:

\[ a^2=b^2+c^2, \]\[ e=\frac ca, \]

and:

\[ PF_1+PF_2=2a. \]

Completion target: Score at least 8/10 on the lesson quiz and independently correct every missed or guessed answer.