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Conic Section Equation

Mathematics · FE Reference Handbook section

Mathematics
7 formulas
10 exam-style examples
~59 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • The general form of the conic section equation is
  • where not both A and C are zero.
  • is the normal form of the conic section equation, if that conic section has a principal axis parallel to a coordinate axis.

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
Parabola vertex form — solve for y-value — Conic Section Equation

A mathematics problem uses Parabola vertex form. Given coefficient (a) = 2.0000; x-value (x) = 3.0000; vertex x (h) = 0.0000; vertex y (k) = -1.0000, determine the y-value (y).

Given

  • coefficient(a)=2.0000coefficient (a) = 2.0000
  • x−value(x)=3.0000x-value (x) = 3.0000
  • vertexx(h)=0.0000vertex x (h) = 0.0000
  • vertexy(k)=−1.0000vertex y (k) = -1.0000

Find

y-value (y)

Start with the thinking

  • The governing relation printed in this handbook section is Parabola vertex form.
  • Everything except y is given, so isolate y 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.
  • Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=a(x−h)2+ky = a (x - h)^2 + k
  2. Step 2 — Rearrange the relation so that y stands alone on the left-hand side.

  3. Step 3

    Listthegivens:coefficient(a)=2.0000,x−value(x)=3.0000,vertexx(h)=0.0000,vertexy(k)=−1.0000List the givens: coefficient (a) = 2.0000, x-value (x) = 3.0000, vertex x (h) = 0.0000, vertex y (k) = -1.0000
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    y=17.0000y = 17.0000
  6. Step 6 — Check: returning y = 17.0000 to

    y=a(x−h)2+ky = a (x - h)^2 + k

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

Answer:
y=17.0000y = 17.0000

Why the other options are there

  • 34.0000 — kept a factor of two that cancels in the correct rearrangement.
  • 8.5000 — dropped that same factor in the other direction.
  • 18.7000 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Conic Section Equation

Example 2
Area of an ellipse — solve for area — Conic Section Equation (2)

an elliptical culvert opening Given semi-major axis (a) = 2.3000 m; semi-minor axis (b) = 1.8000 m, determine the area (A) in m^2.

Given

  • semi−majoraxis(a)=2.3000msemi-major axis (a) = 2.3000 m
  • semi−minoraxis(b)=1.8000msemi-minor axis (b) = 1.8000 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)=2.3000m,semi−minoraxis(b)=1.8000mList the givens: semi-major axis (a) = 2.3000 m, semi-minor axis (b) = 1.8000 m
  4. Step 4 — Substitute the given values:

    A=A=π2.30001.8000A = A = \pi 2.3000 1.8000
  5. Step 5 — Evaluate:

    A = 13.0062\ \text{m^2}
  6. Step 6 — Check: returning A = 13.0062 m^2 to

    A=πabA = \pi a b

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

Answer:
A = 13.0062\ \text{m^2}

Why the other options are there

  • 26.0124 — kept a factor of two that cancels in the correct rearrangement.
  • 6.5031 — dropped that same factor in the other direction.
  • 14.3068 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 3
Eccentricity of an ellipse — solve for eccentricity — Conic Section Equation (3)

Eccentricity measures how far the conic departs from a circle. Given semi-major axis (a) = 6.8000 m; semi-minor axis (b) = 4.4000 m, determine the eccentricity (e).

Given

  • semi−majoraxis(a)=6.8000msemi-major axis (a) = 6.8000 m
  • semi−minoraxis(b)=4.4000msemi-minor axis (b) = 4.4000 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)=6.8000m,semi−minoraxis(b)=4.4000mList the givens: semi-major axis (a) = 6.8000 m, semi-minor axis (b) = 4.4000 m
  4. Step 4 — Substitute the given values:

    e=e=6.80002−4.400026.8000e = e = \dfrac{\sqrt{6.8000^2 - 4.4000^2}}{6.8000}
  5. Step 5 — Evaluate:

    e=0.7624e = 0.7624
  6. Step 6 — Check: returning e = 0.7624 to

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

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

Answer:
e=0.7624e = 0.7624

Why the other options are there

  • 1.5249 — kept a factor of two that cancels in the correct rearrangement.
  • 0.3812 — dropped that same factor in the other direction.
  • 0.8387 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 4
Parabola (vertex form) — solve for elevation — Conic Section Equation (4)

A vertical curve profile is modeled as a parabola. Given curvature coefficient (a) = 0.1000 1/m; station offset (x) = 2.4000 m; vertex station (h) = 0.4000 m; vertex elevation (k) = 13.3000 m, determine the elevation (y) in m.

Given

  • curvaturecoefficient(a)=0.10001/mcurvature coefficient (a) = 0.1000 1/m
  • stationoffset(x)=2.4000mstation offset (x) = 2.4000 m
  • vertexstation(h)=0.4000mvertex station (h) = 0.4000 m
  • vertexelevation(k)=13.3000mvertex elevation (k) = 13.3000 m

Find

elevation (y), in m

Start with the thinking

  • The governing relation printed in this handbook section is Parabola (vertex form).
  • Everything except y is given, so isolate y 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 vertical curve profile is modeled as a parabola.

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=a(x−h)2+ky = a(x - h)^2 + k
  2. Step 2 — Rearrange symbolically for y:

    y=y=a(x−h)2+ky = y = a(x - h)^2 + k
  3. Step 3 — List the givens: curvature coefficient (a) = 0.1000 1/m, station offset (x) = 2.4000 m, vertex station (h) = 0.4000 m, vertex elevation (k) = 13.3000 m.

  4. Step 4 — Substitute the given values:

    y=y=0.1000(2.4000−0.4000)2+13.3000y = y = 0.1000(2.4000 - 0.4000)^2 + 13.3000
  5. Step 5 — Evaluate:

    y=13.7000 my = 13.7000\ \text{m}
  6. Step 6 — Check: returning y = 13.7000 m to

    y=a(x−h)2+ky = a(x - h)^2 + k

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

Answer:
y=13.7000 my = 13.7000\ \text{m}

Why the other options are there

  • 27.4000 — kept a factor of two that cancels in the correct rearrangement.
  • 6.8500 — dropped that same factor in the other direction.
  • 15.0700 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Parabola)

Example 5
Conic section equation — parabola form — solve for y-value — Conic Section Equation (5)

A student evaluates a conic section equation at a given x-value. Given quadratic coefficient (a) = 2.9000; linear coefficient (b) = -0.5000; constant term (c) = 0.1000; x-value (x) = 0.9000, determine the y-value (y).

Given

  • quadraticcoefficient(a)=2.9000quadratic coefficient (a) = 2.9000
  • linearcoefficient(b)=−0.5000linear coefficient (b) = -0.5000
  • constantterm(c)=0.1000constant term (c) = 0.1000
  • x−value(x)=0.9000x-value (x) = 0.9000

Find

y-value (y)

Start with the thinking

  • The governing relation printed in this handbook section is Conic section equation — parabola form.
  • Everything except y is given, so isolate y 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 general conic section equation for a parabola relates its coefficients to a point on the curve.
(0, 0)Conic section (parabola)

Figure 5 — schematic for Conic section equation — parabola form — solve for y-value — Conic Section Equation (5)

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=ax2+bx+cy = a x^2 + b x + c
  2. Step 2 — Rearrange symbolically for y:

    y=ax2+bx+cy = a x^2 + b x + c
  3. Step 3 — List the givens: quadratic coefficient (a) = 2.9000, linear coefficient (b) = -0.5000, constant term (c) = 0.1000, x-value (x) = 0.9000.

  4. Step 4 — Substitute the given values:

    y=2.90000.90002+−0.50000.9000+0.1000y = 2.9000 0.9000^2 + -0.5000 0.9000 + 0.1000
  5. Step 5 — Evaluate:

    y=1.9990y = 1.9990
  6. Step 6 — Check: returning y = 1.9990 to

    y=ax2+bx+cy = a x^2 + b x + c

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

Answer:
y=1.9990y = 1.9990

Why the other options are there

  • 3.9980 — kept a factor of two that cancels in the correct rearrangement.
  • 0.9995 — dropped that same factor in the other direction.
  • 2.1989 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Section Equation

Example 6
Parabola vertex form — solve for coefficient — Conic Section Equation (6)

A mathematics problem uses Parabola vertex form. Given x-value (x) = 5.0000; vertex x (h) = 1.0000; vertex y (k) = 5.0000; y-value (y) = -36.0000, determine the coefficient (a).

Given

  • x−value(x)=5.0000x-value (x) = 5.0000
  • vertexx(h)=1.0000vertex x (h) = 1.0000
  • vertexy(k)=5.0000vertex y (k) = 5.0000
  • y−value(y)=−36.0000y-value (y) = -36.0000

Find

coefficient (a)

Start with the thinking

  • The governing relation printed in this handbook section is Parabola vertex form.
  • 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.
  • Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=a(x−h)2+ky = a (x - h)^2 + k
  2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

  3. Step 3

    Listthegivens:x−value(x)=5.0000,vertexx(h)=1.0000,vertexy(k)=5.0000,y−value(y)=−36.0000List the givens: x-value (x) = 5.0000, vertex x (h) = 1.0000, vertex y (k) = 5.0000, y-value (y) = -36.0000
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    a=−2.5625a = -2.5625
  6. Step 6 — Check: returning a = -2.5625 to

    y=a(x−h)2+ky = a (x - h)^2 + k

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

Answer:
a=−2.5625a = -2.5625

Why the other options are there

  • -5.1250 — kept a factor of two that cancels in the correct rearrangement.
  • -1.2813 — dropped that same factor in the other direction.
  • -2.8188 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Conic Section Equation

Example 7
Area of an ellipse — solve for semi-major axis — Conic Section Equation (7)

an elliptical bearing plate Given area (A) = 7.2000 m^2; semi-minor axis (b) = 1.5000 m, determine the semi-major axis (a) in m.

Given

  • area(A)=7.2000m2area (A) = 7.2000 m^2
  • semi−minoraxis(b)=1.5000msemi-minor axis (b) = 1.5000 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)=7.2000m2,semi−minoraxis(b)=1.5000mList the givens: area (A) = 7.2000 m^2, semi-minor axis (b) = 1.5000 m
  4. Step 4 — Substitute the given values:

    a=a=7.2000π1.5000a = a = \dfrac{7.2000}{\pi 1.5000}
  5. Step 5 — Evaluate:

    a=1.5279 ma = 1.5279\ \text{m}
  6. Step 6 — Check: returning a = 1.5279 m to

    A=πabA = \pi a b

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

Answer:
a=1.5279 ma = 1.5279\ \text{m}

Why the other options are there

  • 3.0558 — kept a factor of two that cancels in the correct rearrangement.
  • 0.7639 — dropped that same factor in the other direction.
  • 1.6807 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 8
Eccentricity of an ellipse — solve for semi-minor axis — Conic Section Equation (8)

Eccentricity measures how far the conic departs from a circle. Given eccentricity (e) = 0.8000; semi-major axis (a) = 2.6000 m, determine the semi-minor axis (b) in m.

Given

  • eccentricity(e)=0.8000eccentricity (e) = 0.8000
  • semi−majoraxis(a)=2.6000msemi-major axis (a) = 2.6000 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)=0.8000,semi−majoraxis(a)=2.6000mList the givens: eccentricity (e) = 0.8000, semi-major axis (a) = 2.6000 m
  4. Step 4 — Substitute the given values:

    b=b=2.60001−0.80002b = b = 2.6000\sqrt{1 - 0.8000^2}
  5. Step 5 — Evaluate:

    b=1.5600 mb = 1.5600\ \text{m}
  6. Step 6 — Check: returning b = 1.5600 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=1.5600 mb = 1.5600\ \text{m}

Why the other options are there

  • 3.1200 — kept a factor of two that cancels in the correct rearrangement.
  • 0.7800 — dropped that same factor in the other direction.
  • 1.7160 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 9
Parabola (vertex form) — solve for curvature coefficient — Conic Section Equation (9)

A vertical curve profile is modeled as a parabola. Given elevation (y) = 11.2000 m; station offset (x) = 4.6000 m; vertex station (h) = 1.8000 m; vertex elevation (k) = 3.8000 m, determine the curvature coefficient (a) in 1/m.

Given

  • elevation(y)=11.2000melevation (y) = 11.2000 m
  • stationoffset(x)=4.6000mstation offset (x) = 4.6000 m
  • vertexstation(h)=1.8000mvertex station (h) = 1.8000 m
  • vertexelevation(k)=3.8000mvertex elevation (k) = 3.8000 m

Find

curvature coefficient (a), in 1/m

Start with the thinking

  • The governing relation printed in this handbook section is Parabola (vertex form).
  • 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 vertical curve profile is modeled as a parabola.

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=a(x−h)2+ky = a(x - h)^2 + k
  2. Step 2 — Rearrange symbolically for a:

    a=a=y−k(x−h)2a = a = \dfrac{y - k}{(x - h)^2}
  3. Step 3 — List the givens: elevation (y) = 11.2000 m, station offset (x) = 4.6000 m, vertex station (h) = 1.8000 m, vertex elevation (k) = 3.8000 m.

  4. Step 4 — Substitute the given values:

    a=a=11.2000−3.8000(4.6000−1.8000)2a = a = \dfrac{11.2000 - 3.8000}{(4.6000 - 1.8000)^2}
  5. Step 5 — Evaluate:

    a=0.9439 1/ma = 0.9439\ \text{1/m}
  6. Step 6 — Check: returning a = 0.9439 1/m to

    y=a(x−h)2+ky = a(x - h)^2 + k

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

Answer:
a=0.9439 1/ma = 0.9439\ \text{1/m}

Why the other options are there

  • 1.8878 — kept a factor of two that cancels in the correct rearrangement.
  • 0.4719 — dropped that same factor in the other direction.
  • 1.0383 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Parabola)

Example 10
Conic section equation — parabola form — solve for constant term — Conic Section Equation (10)

A designer fits a conic section equation to a vertical curve. Given quadratic coefficient (a) = 3.0000; linear coefficient (b) = -3.4000; x-value (x) = -1.9000; y-value (y) = -10.1000, determine the constant term (c).

Given

  • quadraticcoefficient(a)=3.0000quadratic coefficient (a) = 3.0000
  • linearcoefficient(b)=−3.4000linear coefficient (b) = -3.4000
  • x−value(x)=−1.9000x-value (x) = -1.9000
  • y−value(y)=−10.1000y-value (y) = -10.1000

Find

constant term (c)

Start with the thinking

  • The governing relation printed in this handbook section is Conic section equation — parabola form.
  • Everything except c is given, so isolate c 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 general conic section equation for a parabola relates its coefficients to a point on the curve.
(0, 0)Conic section (parabola)

Figure 10 — schematic for Conic section equation — parabola form — solve for constant term — Conic Section Equation (10)

Step-by-step solution

  1. Step 1 — State the governing relation:

    y=ax2+bx+cy = a x^2 + b x + c
  2. Step 2 — Rearrange symbolically for c:

    c=y−ax2−bxc = y - a x^2 - b x
  3. Step 3 — List the givens: quadratic coefficient (a) = 3.0000, linear coefficient (b) = -3.4000, x-value (x) = -1.9000, y-value (y) = -10.1000.

  4. Step 4 — Substitute the given values:

    c=−10.1000−3.0000−1.90002−−3.4000−1.9000c = -10.1000 - 3.0000 -1.9000^2 - -3.4000 -1.9000
  5. Step 5 — Evaluate:

    c=−27.3900c = -27.3900
  6. Step 6 — Check: returning c = -27.3900 to

    y=ax2+bx+cy = a x^2 + b x + c

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

Answer:
c=−27.3900c = -27.3900

Why the other options are there

  • -54.7800 — kept a factor of two that cancels in the correct rearrangement.
  • -13.6950 — dropped that same factor in the other direction.
  • -30.1290 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Section Equation

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