Conic Sections

Example 1

Parabola vertex form — solve for y-value — Conic Sections

A mathematics problem uses Parabola vertex form. Given coefficient (a) = 1.7000; x-value (x) = 4.5000; vertex x (h) = -3.0000; vertex y (k) = -2.5000, determine the y-value (y).

Given

$coefficient (a) = 1.7000$
$x-value (x) = 4.5000$
$vertex x (h) = -3.0000$
$vertex y (k) = -2.5000$

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

Step 1 — State the governing relation:
$y = a (x - h)^2 + k$
Step 2 — Rearrange the relation so that y stands alone on the left-hand side.
Step 3 — List the givens: coefficient (a) = 1.7000, x-value (x) = 4.5000, vertex x (h) = -3.0000, vertex y (k) = -2.5000.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$y = 93.1250$
Step 6 — Check: returning y = 93.1250 to
$y = a (x - h)^2 + k$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = 93.1250

Why the other options are there

186.3 — kept a factor of two that cancels in the correct rearrangement.
46.5625 — dropped that same factor in the other direction.
102.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Conic Sections

Example 2

Area of an ellipse — solve for area — Conic Sections (2)

an elliptical culvert opening Given semi-major axis (a) = 2.1000 m; semi-minor axis (b) = 0.7000 m, determine the area (A) in m^2.

Given

$semi-major axis (a) = 2.1000 m$
$semi-minor axis (b) = 0.7000 m$

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

Step 1 — State the governing relation:
$A = \pi a b$
Step 2 — Rearrange symbolically for A:
$A = A = \pi a b$
Step 3
$List the givens: semi-major axis (a) = 2.1000 m, semi-minor axis (b) = 0.7000 m$
Step 4 — Substitute the given values:
$A = A = \pi 2.1000 0.7000$
Step 5 — Evaluate:
$A = 4.6181\ \text{m^2}$
Step 6 — Check: returning A = 4.6181 m^2 to
$A = \pi a b$
reproduces the given quantities, and both sides carry the same units.

Answer:

A = 4.6181\ \text{m^2}

Why the other options are there

9.2363 — kept a factor of two that cancels in the correct rearrangement.
2.3091 — dropped that same factor in the other direction.
5.0800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 3

Eccentricity of an ellipse — solve for eccentricity — Conic Sections (3)

Eccentricity measures how far the conic departs from a circle. Given semi-major axis (a) = 3.6000 m; semi-minor axis (b) = 1.2000 m, determine the eccentricity (e).

Given

$semi-major axis (a) = 3.6000 m$
$semi-minor axis (b) = 1.2000 m$

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

Step 1 — State the governing relation:
$e = \dfrac{\sqrt{a^2 - b^2}}{a}$
Step 2 — Rearrange symbolically for e:
$e = e = \dfrac{\sqrt{a^2 - b^2}}{a}$
Step 3
$List the givens: semi-major axis (a) = 3.6000 m, semi-minor axis (b) = 1.2000 m$
Step 4 — Substitute the given values:
$e = e = \dfrac{\sqrt{3.6000^2 - 1.2000^2}}{3.6000}$
Step 5 — Evaluate:
$e = 0.9428$
Step 6 — Check: returning e = 0.9428 to
$e = \dfrac{\sqrt{a^2 - b^2}}{a}$
reproduces the given quantities, and both sides carry the same units.

Answer:

e = 0.9428

Why the other options are there

1.8856 — kept a factor of two that cancels in the correct rearrangement.
0.4714 — dropped that same factor in the other direction.
1.0371 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 4

Parabola (vertex form) — solve for elevation — Conic Sections (4)

A vertical curve profile is modeled as a parabola. Given curvature coefficient (a) = 0.2000 1/m; station offset (x) = 18.0000 m; vertex station (h) = 3.1000 m; vertex elevation (k) = 7.7000 m, determine the elevation (y) in m.

Given

$curvature coefficient (a) = 0.2000 1/m$
$station offset (x) = 18.0000 m$
$vertex station (h) = 3.1000 m$
$vertex elevation (k) = 7.7000 m$

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

Step 1 — State the governing relation:
$y = a(x - h)^2 + k$
Step 2 — Rearrange symbolically for y:
$y = y = a(x - h)^2 + k$
Step 3 — List the givens: curvature coefficient (a) = 0.2000 1/m, station offset (x) = 18.0000 m, vertex station (h) = 3.1000 m, vertex elevation (k) = 7.7000 m.
Step 4 — Substitute the given values:
$y = y = 0.2000(18.0000 - 3.1000)^2 + 7.7000$
Step 5 — Evaluate:
$y = 52.1020\ \text{m}$
Step 6 — Check: returning y = 52.1020 m to
$y = a(x - h)^2 + k$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = 52.1020\ \text{m}

Why the other options are there

104.2 — kept a factor of two that cancels in the correct rearrangement.
26.0510 — dropped that same factor in the other direction.
57.3122 — 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 Sections (5)

A student evaluates a conic section equation at a given x-value. Given quadratic coefficient (a) = 0.6000; linear coefficient (b) = -3.5000; constant term (c) = -2.6000; x-value (x) = 1.8000, determine the y-value (y).

Given

$quadratic coefficient (a) = 0.6000$
$linear coefficient (b) = -3.5000$
$constant term (c) = -2.6000$
$x-value (x) = 1.8000$

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.

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

Step-by-step solution

Step 1 — State the governing relation:
$y = a x^2 + b x + c$
Step 2 — Rearrange symbolically for y:
$y = a x^2 + b x + c$
Step 3 — List the givens: quadratic coefficient (a) = 0.6000, linear coefficient (b) = -3.5000, constant term (c) = -2.6000, x-value (x) = 1.8000.
Step 4 — Substitute the given values:
$y = 0.6000 1.8000^2 + -3.5000 1.8000 + -2.6000$
Step 5 — Evaluate:
$y = -6.9560$
Step 6 — Check: returning y = -6.9560 to
$y = a x^2 + b x + c$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = -6.9560

Why the other options are there

-13.9120 — kept a factor of two that cancels in the correct rearrangement.
-3.4780 — dropped that same factor in the other direction.
-7.6516 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Section Equation

Example 6

Parabola vertex form — solve for coefficient — Conic Sections (6)

A mathematics problem uses Parabola vertex form. Given x-value (x) = -1.5000; vertex x (h) = 2.0000; vertex y (k) = -5.0000; y-value (y) = -48.3000, determine the coefficient (a).

Given

$x-value (x) = -1.5000$
$vertex x (h) = 2.0000$
$vertex y (k) = -5.0000$
$y-value (y) = -48.3000$

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

Step 1 — State the governing relation:
$y = a (x - h)^2 + k$
Step 2 — Rearrange the relation so that a stands alone on the left-hand side.
Step 3
$List the givens: x-value (x) = -1.5000, vertex x (h) = 2.0000, vertex y (k) = -5.0000, y-value (y) = -48.3000$
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$a = -3.5347$
Step 6 — Check: returning a = -3.5347 to
$y = a (x - h)^2 + k$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = -3.5347

Why the other options are there

-7.0694 — kept a factor of two that cancels in the correct rearrangement.
-1.7673 — dropped that same factor in the other direction.
-3.8882 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Conic Sections

Example 7

Area of an ellipse — solve for semi-major axis — Conic Sections (7)

an elliptical bearing plate Given area (A) = 23.4000 m^2; semi-minor axis (b) = 2.3000 m, determine the semi-major axis (a) in m.

Given

$area (A) = 23.4000 m^2$
$semi-minor axis (b) = 2.3000 m$

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

Step 1 — State the governing relation:
$A = \pi a b$
Step 2 — Rearrange symbolically for a:
$a = a = \dfrac{A}{\pi b}$
Step 3
$List the givens: area (A) = 23.4000 m^2, semi-minor axis (b) = 2.3000 m$
Step 4 — Substitute the given values:
$a = a = \dfrac{23.4000}{\pi 2.3000}$
Step 5 — Evaluate:
$a = 3.2385\ \text{m}$
Step 6 — Check: returning a = 3.2385 m to
$A = \pi a b$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = 3.2385\ \text{m}

Why the other options are there

6.4769 — kept a factor of two that cancels in the correct rearrangement.
1.6192 — dropped that same factor in the other direction.
3.5623 — 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 Sections (8)

Eccentricity measures how far the conic departs from a circle. Given eccentricity (e) = 0.1000; semi-major axis (a) = 3.1000 m, determine the semi-minor axis (b) in m.

Given

$eccentricity (e) = 0.1000$
$semi-major axis (a) = 3.1000 m$

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

Step 1 — State the governing relation:
$e = \dfrac{\sqrt{a^2 - b^2}}{a}$
Step 2 — Rearrange symbolically for b:
$b = b = a\sqrt{1 - e^2}$
Step 3
$List the givens: eccentricity (e) = 0.1000, semi-major axis (a) = 3.1000 m$
Step 4 — Substitute the given values:
$b = b = 3.1000\sqrt{1 - 0.1000^2}$
Step 5 — Evaluate:
$b = 3.0845\ \text{m}$
Step 6 — Check: returning b = 3.0845 m to
$e = \dfrac{\sqrt{a^2 - b^2}}{a}$
reproduces the given quantities, and both sides carry the same units.

Answer:

b = 3.0845\ \text{m}

Why the other options are there

6.1689 — kept a factor of two that cancels in the correct rearrangement.
1.5422 — dropped that same factor in the other direction.
3.3929 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Sections (Ellipse)

Example 9

Parabola (vertex form) — solve for curvature coefficient — Conic Sections (9)

A vertical curve profile is modeled as a parabola. Given elevation (y) = 23.0000 m; station offset (x) = 27.2000 m; vertex station (h) = 9.3000 m; vertex elevation (k) = 3.9000 m, determine the curvature coefficient (a) in 1/m.

Given

$elevation (y) = 23.0000 m$
$station offset (x) = 27.2000 m$
$vertex station (h) = 9.3000 m$
$vertex elevation (k) = 3.9000 m$

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

Step 1 — State the governing relation:
$y = a(x - h)^2 + k$
Step 2 — Rearrange symbolically for a:
$a = a = \dfrac{y - k}{(x - h)^2}$
Step 3 — List the givens: elevation (y) = 23.0000 m, station offset (x) = 27.2000 m, vertex station (h) = 9.3000 m, vertex elevation (k) = 3.9000 m.
Step 4 — Substitute the given values:
$a = a = \dfrac{23.0000 - 3.9000}{(27.2000 - 9.3000)^2}$
Step 5 — Evaluate:
$a = 0.0596\ \text{1/m}$
Step 6 — Check: returning a = 0.0596 1/m to
$y = a(x - h)^2 + k$
reproduces the given quantities, and both sides carry the same units.

Answer:

a = 0.0596\ \text{1/m}

Why the other options are there

0.1192 — kept a factor of two that cancels in the correct rearrangement.
0.0298 — dropped that same factor in the other direction.
0.0656 — 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 Sections (10)

A designer fits a conic section equation to a vertical curve. Given quadratic coefficient (a) = 3.0000; linear coefficient (b) = 3.6000; x-value (x) = 3.6000; y-value (y) = 59.8000, determine the constant term (c).

Given

$quadratic coefficient (a) = 3.0000$
$linear coefficient (b) = 3.6000$
$x-value (x) = 3.6000$
$y-value (y) = 59.8000$

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.

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

Step-by-step solution

Step 1 — State the governing relation:
$y = a x^2 + b x + c$
Step 2 — Rearrange symbolically for c:
$c = y - a x^2 - b x$
Step 3 — List the givens: quadratic coefficient (a) = 3.0000, linear coefficient (b) = 3.6000, x-value (x) = 3.6000, y-value (y) = 59.8000.
Step 4 — Substitute the given values:
$c = 59.8000 - 3.0000 3.6000^2 - 3.6000 3.6000$
Step 5 — Evaluate:
$c = 7.9600$
Step 6 — Check: returning c = 7.9600 to
$y = a x^2 + b x + c$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 7.9600

Why the other options are there

15.9200 — kept a factor of two that cancels in the correct rearrangement.
3.9800 — dropped that same factor in the other direction.
8.7560 — rounded an intermediate value before the final step.

Reference: FE Handbook — Conic Section Equation

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Conic Sections

Handbook notes for this section

Core formulas for this FE topic

e=eccentricity=cos⁡θ/(cos⁡ϕ)e = eccentricity = \cos \theta/(\cos \phi)e=eccentricity=cosθ/(cosϕ)

Case1.Parabolae=1:Case 1. Parabola e = 1:Case1.Parabolae=1:

(y−k)2=2p(x−h);Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,(y - k)2 = 2p(x - h); Center at (h, k) is the standard form of the equation. When h = k = 0,(y−k)2=2p(x−h);Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,

Focus:(p/2,0);Directrix:x=−p/2Focus: (p/2, 0); Directrix: x = -p/2Focus:(p/2,0);Directrix:x=−p/2

+=1;Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,+ = 1; Center at (h, k) is the standard form of the equation. When h = k = 0,+=1;Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,

Eccentricity:e=1−b2/a2i=c/aEccentricity: e = 1 - _b 2/a 2i = c/aEccentricity:e=1−b​2/a2i=c/a

b=a1−e2;b = a 1 - e 2;b=a1−e2;

Focus:(±ae,0);Directrix:x=±a/eFocus: (\pm ae, 0); Directrix: x = \pm a/eFocus:(±ae,0);Directrix:x=±a/e

x−hhy−ki^ x - hh _ y - kix−hhy​−ki

−=1;- = 1;−=1;

Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,Center at (h, k) is the standard form of the equation. When h = k = 0,Centerat(h,k)isthestandardformoftheequation.Whenh=k=0,

Eccentricity:e=1+b2/a2i=c/aEccentricity: e = 1 + _b 2/a 2i = c/aEccentricity:e=1+b​2/a2i=c/a

b=ae2−1;b = a e 2 - 1;b=ae2−1;

Focus:(±ae,0);Directrix:x=±a/eFocus: (\pm ae, 0); Directrix: x = \pm a/eFocus:(±ae,0);Directrix:x=±a/e

Case4.Circlee=0:Case 4. Circle e = 0:Case4.Circlee=0:

(x−h)2+(y−k)2=r2;Centerat(h,k)isthestandardformoftheequationwithradius(x - h)2 + (y - k)2 = r_{2}; Center at (h, k) is the standard form of the equation with radius(x−h)2+(y−k)2=r2​;Centerat(h,k)isthestandardformoftheequationwithradius

x−hh+y−ki^ x - hh + _ y - k ix−hh+y​−ki

t2=(x′−h)2+(y′−k)2−r2t_{2} = (x' - h)2 + (y' - k)2 - r_{2}t2​=(x′−h)2+(y′−k)2−r2​

Worked exam-style examples