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Ellipse

Mathematics · FE Reference Handbook section

Mathematics
5 formulas
10 exam-style examples
~55 min
All Mathematics lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Area of an ellipse — solve for area — Ellipse

an elliptical manhole cover Given semi-major axis (a) = 5.9000 m; semi-minor axis (b) = 2.0000 m, determine the area (A) in m^2.

Given

  • semi−majoraxis(a)=5.9000msemi-major axis (a) = 5.9000 m
  • semi−minoraxis(b)=2.0000msemi-minor axis (b) = 2.0000 m

Find

area (A), in m^2

Start with the thinking

  • The governing relation printed in this handbook section is Area of an ellipse.
  • Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A survey monument plate is machined as an ellipse.

Step-by-step solution

  1. Step 1 — State the governing relation:

    A=πabA = \pi a b
  2. Step 2 — Rearrange symbolically for A:

    A=A=πabA = A = \pi a b
  3. Step 3

    Listthegivens:semi−majoraxis(a)=5.9000m,semi−minoraxis(b)=2.0000mList the givens: semi-major axis (a) = 5.9000 m, semi-minor axis (b) = 2.0000 m
  4. Step 4 — Substitute the given values:

    A=A=π5.90002.0000A = A = \pi 5.9000 2.0000
  5. Step 5 — Evaluate:

    A = 37.0708\ \text{m^2}
  6. Step 6 — Check: returning A = 37.0708 m^2 to

    A=πabA = \pi a b

    reproduces the given quantities, and both sides carry the same units.

Answer:
A = 37.0708\ \text{m^2}

Why the other options are there

  • 74.1416 — kept a factor of two that cancels in the correct rearrangement.
  • 18.5354 — dropped that same factor in the other direction.
  • 40.7779 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 2
Eccentricity of an ellipse — solve for eccentricity — Ellipse (2)

Eccentricity measures how far the conic departs from a circle. Given semi-major axis (a) = 3.9000 m; semi-minor axis (b) = 2.6000 m, determine the eccentricity (e).

Given

  • semi−majoraxis(a)=3.9000msemi-major axis (a) = 3.9000 m
  • semi−minoraxis(b)=2.6000msemi-minor axis (b) = 2.6000 m

Find

eccentricity (e)

Start with the thinking

  • The governing relation printed in this handbook section is Eccentricity of an ellipse.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Eccentricity measures how far the conic departs from a circle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for e:

    e=e=a2−b2ae = e = \dfrac{\sqrt{a^2 - b^2}}{a}
  3. Step 3

    Listthegivens:semi−majoraxis(a)=3.9000m,semi−minoraxis(b)=2.6000mList the givens: semi-major axis (a) = 3.9000 m, semi-minor axis (b) = 2.6000 m
  4. Step 4 — Substitute the given values:

    e=e=3.90002−2.600023.9000e = e = \dfrac{\sqrt{3.9000^2 - 2.6000^2}}{3.9000}
  5. Step 5 — Evaluate:

    e=0.7454e = 0.7454
  6. Step 6 — Check: returning e = 0.7454 to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=0.7454e = 0.7454

Why the other options are there

  • 1.4907 — kept a factor of two that cancels in the correct rearrangement.
  • 0.3727 — dropped that same factor in the other direction.
  • 0.8199 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 3
Ellipse — semi-axes and eccentricity — solve for eccentricity — Ellipse (3)

A student computes the eccentricity of an ellipse from its axes. Given semi-major axis (a) = 19.5000 ft; semi-minor axis (b) = 6.0000 ft, determine the eccentricity (e).

Given

  • semi−majoraxis(a)=19.5000ftsemi-major axis (a) = 19.5000 ft
  • semi−minoraxis(b)=6.0000ftsemi-minor axis (b) = 6.0000 ft

Find

eccentricity (e)

Start with the thinking

  • The governing relation printed in this handbook section is Ellipse — semi-axes and eccentricity.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The ellipse relation ties the semi-major axis, semi-minor axis, and eccentricity of the conic.
(0, 0)Ellipse geometry

Figure 3 — schematic for Ellipse — semi-axes and eccentricity — solve for eccentricity — Ellipse (3)

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for e:

    e=a2−b2ae = \dfrac{\sqrt{a^2-b^2}}{a}
  3. Step 3

    Listthegivens:semi−majoraxis(a)=19.5000ft,semi−minoraxis(b)=6.0000ftList the givens: semi-major axis (a) = 19.5000 ft, semi-minor axis (b) = 6.0000 ft
  4. Step 4 — Substitute the given values:

    e=19.50002−6.0000219.5000e = \dfrac{\sqrt{19.5000^2-6.0000^2}}{19.5000}
  5. Step 5 — Evaluate:

    e=0.9515e = 0.9515
  6. Step 6 — Check: returning e = 0.9515 to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=0.9515e = 0.9515

Why the other options are there

  • 1.9030 — kept a factor of two that cancels in the correct rearrangement.
  • 0.4757 — dropped that same factor in the other direction.
  • 1.0466 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ellipse

Example 4
Area of an ellipse — solve for semi-major axis — Ellipse (4)

an elliptical culvert opening Given area (A) = 25.1000 m^2; semi-minor axis (b) = 2.3000 m, determine the semi-major axis (a) in m.

Given

  • area(A)=25.1000m2area (A) = 25.1000 m^2
  • semi−minoraxis(b)=2.3000msemi-minor axis (b) = 2.3000 m

Find

semi-major axis (a), in m

Start with the thinking

  • The governing relation printed in this handbook section is Area of an ellipse.
  • Everything except a is given, so isolate a symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A survey monument plate is machined as an ellipse.

Step-by-step solution

  1. Step 1 — State the governing relation:

    A=πabA = \pi a b
  2. Step 2 — Rearrange symbolically for a:

    a=a=Aπba = a = \dfrac{A}{\pi b}
  3. Step 3

    Listthegivens:area(A)=25.1000m2,semi−minoraxis(b)=2.3000mList the givens: area (A) = 25.1000 m^2, semi-minor axis (b) = 2.3000 m
  4. Step 4 — Substitute the given values:

    a=a=25.1000π2.3000a = a = \dfrac{25.1000}{\pi 2.3000}
  5. Step 5 — Evaluate:

    a=3.4737 ma = 3.4737\ \text{m}
  6. Step 6 — Check: returning a = 3.4737 m to

    A=πabA = \pi a b

    reproduces the given quantities, and both sides carry the same units.

Answer:
a=3.4737 ma = 3.4737\ \text{m}

Why the other options are there

  • 6.9475 — kept a factor of two that cancels in the correct rearrangement.
  • 1.7369 — dropped that same factor in the other direction.
  • 3.8211 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 5
Eccentricity of an ellipse — solve for semi-minor axis — Ellipse (5)

Eccentricity measures how far the conic departs from a circle. Given eccentricity (e) = 0.9000; semi-major axis (a) = 2.7000 m, determine the semi-minor axis (b) in m.

Given

  • eccentricity(e)=0.9000eccentricity (e) = 0.9000
  • semi−majoraxis(a)=2.7000msemi-major axis (a) = 2.7000 m

Find

semi-minor axis (b), in m

Start with the thinking

  • The governing relation printed in this handbook section is Eccentricity of an ellipse.
  • Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Eccentricity measures how far the conic departs from a circle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for b:

    b=b=a1−e2b = b = a\sqrt{1 - e^2}
  3. Step 3

    Listthegivens:eccentricity(e)=0.9000,semi−majoraxis(a)=2.7000mList the givens: eccentricity (e) = 0.9000, semi-major axis (a) = 2.7000 m
  4. Step 4 — Substitute the given values:

    b=b=2.70001−0.90002b = b = 2.7000\sqrt{1 - 0.9000^2}
  5. Step 5 — Evaluate:

    b=1.1769 mb = 1.1769\ \text{m}
  6. Step 6 — Check: returning b = 1.1769 m to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
b=1.1769 mb = 1.1769\ \text{m}

Why the other options are there

  • 2.3538 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5885 — dropped that same factor in the other direction.
  • 1.2946 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 6
Ellipse — semi-axes and eccentricity — solve for semi-minor axis — Ellipse (6)

An engineer designs an elliptical arch and checks its eccentricity. Given semi-major axis (a) = 19.0000 ft; eccentricity (e) = 0.9130, determine the semi-minor axis (b) in ft.

Given

  • semi−majoraxis(a)=19.0000ftsemi-major axis (a) = 19.0000 ft
  • eccentricity(e)=0.9130eccentricity (e) = 0.9130

Find

semi-minor axis (b), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Ellipse — semi-axes and eccentricity.
  • Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The ellipse relation ties the semi-major axis, semi-minor axis, and eccentricity of the conic.
(0, 0)Ellipse geometry

Figure 6 — schematic for Ellipse — semi-axes and eccentricity — solve for semi-minor axis — Ellipse (6)

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for b:

    b=a1−e2b = a\sqrt{1-e^2}
  3. Step 3

    Listthegivens:semi−majoraxis(a)=19.0000ft,eccentricity(e)=0.9130List the givens: semi-major axis (a) = 19.0000 ft, eccentricity (e) = 0.9130
  4. Step 4 — Substitute the given values:

    b=19.00001−0.91302b = 19.0000\sqrt{1-0.9130^2}
  5. Step 5 — Evaluate:

    b=7.7512 ftb = 7.7512\ \text{ft}
  6. Step 6 — Check: returning b = 7.7512 ft to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
b=7.7512 ftb = 7.7512\ \text{ft}

Why the other options are there

  • 15.5025 — kept a factor of two that cancels in the correct rearrangement.
  • 3.8756 — dropped that same factor in the other direction.
  • 8.5264 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ellipse

Example 7
Area of an ellipse — solve for semi-minor axis — Ellipse (7)

an elliptical bearing plate Given area (A) = 31.9000 m^2; semi-major axis (a) = 3.2000 m, determine the semi-minor axis (b) in m.

Given

  • area(A)=31.9000m2area (A) = 31.9000 m^2
  • semi−majoraxis(a)=3.2000msemi-major axis (a) = 3.2000 m

Find

semi-minor axis (b), in m

Start with the thinking

  • The governing relation printed in this handbook section is Area of an ellipse.
  • Everything except b is given, so isolate b symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A survey monument plate is machined as an ellipse.

Step-by-step solution

  1. Step 1 — State the governing relation:

    A=πabA = \pi a b
  2. Step 2 — Rearrange symbolically for b:

    b=b=Aπab = b = \dfrac{A}{\pi a}
  3. Step 3

    Listthegivens:area(A)=31.9000m2,semi−majoraxis(a)=3.2000mList the givens: area (A) = 31.9000 m^2, semi-major axis (a) = 3.2000 m
  4. Step 4 — Substitute the given values:

    b=b=31.9000π3.2000b = b = \dfrac{31.9000}{\pi 3.2000}
  5. Step 5 — Evaluate:

    b=3.1732 mb = 3.1732\ \text{m}
  6. Step 6 — Check: returning b = 3.1732 m to

    A=πabA = \pi a b

    reproduces the given quantities, and both sides carry the same units.

Answer:
b=3.1732 mb = 3.1732\ \text{m}

Why the other options are there

  • 6.3463 — kept a factor of two that cancels in the correct rearrangement.
  • 1.5866 — dropped that same factor in the other direction.
  • 3.4905 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 8
Eccentricity of an ellipse — solve for eccentricity (case 2) — Ellipse (8)

Eccentricity measures how far the conic departs from a circle. Given semi-major axis (a) = 5.2000 m; semi-minor axis (b) = 1.0000 m, determine the eccentricity (e).

Given

  • semi−majoraxis(a)=5.2000msemi-major axis (a) = 5.2000 m
  • semi−minoraxis(b)=1.0000msemi-minor axis (b) = 1.0000 m

Find

eccentricity (e)

Start with the thinking

  • The governing relation printed in this handbook section is Eccentricity of an ellipse.
  • Everything except e is given, so isolate e symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Eccentricity measures how far the conic departs from a circle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for e:

    e=e=a2−b2ae = e = \dfrac{\sqrt{a^2 - b^2}}{a}
  3. Step 3

    Listthegivens:semi−majoraxis(a)=5.2000m,semi−minoraxis(b)=1.0000mList the givens: semi-major axis (a) = 5.2000 m, semi-minor axis (b) = 1.0000 m
  4. Step 4 — Substitute the given values:

    e=e=5.20002−1.000025.2000e = e = \dfrac{\sqrt{5.2000^2 - 1.0000^2}}{5.2000}
  5. Step 5 — Evaluate:

    e=0.9813e = 0.9813
  6. Step 6 — Check: returning e = 0.9813 to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
e=0.9813e = 0.9813

Why the other options are there

  • 1.9627 — kept a factor of two that cancels in the correct rearrangement.
  • 0.4907 — dropped that same factor in the other direction.
  • 1.0795 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 9
Ellipse — semi-axes and eccentricity — solve for semi-major axis — Ellipse (9)

A designer lays out an ellipse for a decorative footing outline. Given semi-minor axis (b) = 4.5000 ft; eccentricity (e) = 0.6020, determine the semi-major axis (a) in ft.

Given

  • semi−minoraxis(b)=4.5000ftsemi-minor axis (b) = 4.5000 ft
  • eccentricity(e)=0.6020eccentricity (e) = 0.6020

Find

semi-major axis (a), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Ellipse — semi-axes and eccentricity.
  • Everything except a is given, so isolate a symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The ellipse relation ties the semi-major axis, semi-minor axis, and eccentricity of the conic.
(0, 0)Ellipse geometry

Figure 9 — schematic for Ellipse — semi-axes and eccentricity — solve for semi-major axis — Ellipse (9)

Step-by-step solution

  1. Step 1 — State the governing relation:

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}
  2. Step 2 — Rearrange symbolically for a:

    a=b1−e2a = \dfrac{b}{\sqrt{1-e^2}}
  3. Step 3

    Listthegivens:semi−minoraxis(b)=4.5000ft,eccentricity(e)=0.6020List the givens: semi-minor axis (b) = 4.5000 ft, eccentricity (e) = 0.6020
  4. Step 4 — Substitute the given values:

    a=4.50001−0.60202a = \dfrac{4.5000}{\sqrt{1-0.6020^2}}
  5. Step 5 — Evaluate:

    a=5.6356 fta = 5.6356\ \text{ft}
  6. Step 6 — Check: returning a = 5.6356 ft to

    e=a2−b2ae = \dfrac{\sqrt{a^2 - b^2}}{a}

    reproduces the given quantities, and both sides carry the same units.

Answer:
a=5.6356 fta = 5.6356\ \text{ft}

Why the other options are there

  • 11.2712 — kept a factor of two that cancels in the correct rearrangement.
  • 2.8178 — dropped that same factor in the other direction.
  • 6.1992 — rounded an intermediate value before the final step.

Reference: FE Handbook — Ellipse

Example 10
Area of an ellipse — solve for area (case 2) — Ellipse (10)

an elliptical manhole cover Given semi-major axis (a) = 1.3000 m; semi-minor axis (b) = 2.2000 m, determine the area (A) in m^2.

Given

  • semi−majoraxis(a)=1.3000msemi-major axis (a) = 1.3000 m
  • semi−minoraxis(b)=2.2000msemi-minor axis (b) = 2.2000 m

Find

area (A), in m^2

Start with the thinking

  • The governing relation printed in this handbook section is Area of an ellipse.
  • Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A survey monument plate is machined as an ellipse.

Step-by-step solution

  1. Step 1 — State the governing relation:

    A=πabA = \pi a b
  2. Step 2 — Rearrange symbolically for A:

    A=A=πabA = A = \pi a b
  3. Step 3

    Listthegivens:semi−majoraxis(a)=1.3000m,semi−minoraxis(b)=2.2000mList the givens: semi-major axis (a) = 1.3000 m, semi-minor axis (b) = 2.2000 m
  4. Step 4 — Substitute the given values:

    A=A=π1.30002.2000A = A = \pi 1.3000 2.2000
  5. Step 5 — Evaluate:

    A = 8.9850\ \text{m^2}
  6. Step 6 — Check: returning A = 8.9850 m^2 to

    A=πabA = \pi a b

    reproduces the given quantities, and both sides carry the same units.

Answer:
A = 8.9850\ \text{m^2}

Why the other options are there

  • 17.9699 — kept a factor of two that cancels in the correct rearrangement.
  • 4.4925 — dropped that same factor in the other direction.
  • 9.8835 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

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