College Algebra

Ellipses

Last updated: March 2026 · Advanced
Before you start

You should be comfortable with:

An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (the foci) is a constant. If you have ever drawn an oval by pinning two tacks on a board, looping a string around them, and tracing with a pencil, you have drawn an ellipse. Planets orbit the sun in ellipses, whispering galleries use elliptical ceilings to transmit sound, and elliptical arches appear in architecture worldwide.

The Definition and Key Measurements

For any point PP on the ellipse with foci F1F_1 and F2F_2:

d(P,F1)+d(P,F2)=2ad(P, F_1) + d(P, F_2) = 2a

where aa is the semi-major axis (half the length of the longest diameter).

Key measurements:

  • aa = semi-major axis (vertex to center, along the longer direction)
  • bb = semi-minor axis (co-vertex to center, along the shorter direction)
  • cc = distance from center to each focus

The relationship between them is:

c2=a2b2c^2 = a^2 - b^2

This means a>ba > b always (the semi-major axis is the longer one), and cc is always less than aa (the foci are inside the ellipse).

Standard Forms

Horizontal Major Axis (wider than tall)

(xh)2a2+(yk)2b2=1(a>b)\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \quad (a > b)

  • Center: (h,k)(h, k)
  • Vertices: (h±a,k)(h \pm a, k) — endpoints of the major axis
  • Co-vertices: (h,k±b)(h, k \pm b) — endpoints of the minor axis
  • Foci: (h±c,k)(h \pm c, k) where c=a2b2c = \sqrt{a^2 - b^2}

Vertical Major Axis (taller than wide)

(xh)2b2+(yk)2a2=1(a>b)\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \quad (a > b)

  • Center: (h,k)(h, k)
  • Vertices: (h,k±a)(h, k \pm a) — endpoints of the major axis
  • Co-vertices: (h±b,k)(h \pm b, k) — endpoints of the minor axis
  • Foci: (h,k±c)(h, k \pm c) where c=a2b2c = \sqrt{a^2 - b^2}

Key pattern: The larger denominator is always a2a^2. If it is under the xx-term, the major axis is horizontal. If it is under the yy-term, the major axis is vertical.

Ellipse with Horizontal Major Axis

(-a, 0)(a, 0)(0, b)(0, -b)F1F2abc

Eccentricity

The eccentricity measures how “elongated” the ellipse is:

e=cae = \frac{c}{a}

Since cc is always between 00 and aa, eccentricity satisfies 0e<10 \le e < 1:

  • e=0e = 0: the ellipse is a perfect circle (c=0c = 0, foci at center)
  • ee close to 1: the ellipse is very elongated (foci near the ends)
  • Earth’s orbit: e0.017e \approx 0.017 (nearly circular)
  • Pluto’s orbit: e0.248e \approx 0.248 (noticeably elongated)

Worked Examples

Example 1: Identifying Parts from the Equation

Find the center, vertices, co-vertices, foci, and eccentricity of:

(x1)225+(y+3)29=1\frac{(x - 1)^2}{25} + \frac{(y + 3)^2}{9} = 1

Step 1: Center: (1,3)(1, -3).

Step 2: Since 25>925 > 9, the larger denominator is under xx, so the major axis is horizontal. a2=25a^2 = 25 so a=5a = 5. b2=9b^2 = 9 so b=3b = 3.

Step 3: c2=a2b2=259=16c^2 = a^2 - b^2 = 25 - 9 = 16, so c=4c = 4.

  • Vertices: (1±5,3)=(4,3)(1 \pm 5, -3) = (-4, -3) and (6,3)(6, -3)
  • Co-vertices: (1,3±3)=(1,6)(1, -3 \pm 3) = (1, -6) and (1,0)(1, 0)
  • Foci: (1±4,3)=(3,3)(1 \pm 4, -3) = (-3, -3) and (5,3)(5, -3)
  • Eccentricity: e=45=0.8e = \frac{4}{5} = 0.8

Example 2: Vertical Major Axis

Find the foci of:

x24+y216=1\frac{x^2}{4} + \frac{y^2}{16} = 1

The larger denominator 1616 is under y2y^2, so the major axis is vertical. a2=16a^2 = 16, b2=4b^2 = 4.

c2=164=12,c=23c^2 = 16 - 4 = 12, \quad c = 2\sqrt{3}

Foci: (0,±23)(0,±3.46)(0, \pm 2\sqrt{3}) \approx (0, \pm 3.46)

Example 3: Writing the Equation from Given Information

Write the equation of an ellipse with center (0,0)(0, 0), vertices at (±6,0)(\pm 6, 0), and foci at (±4,0)(\pm 4, 0).

Step 1: Vertices at (±6,0)(\pm 6, 0) means a=6a = 6 and the major axis is horizontal.

Step 2: Foci at (±4,0)(\pm 4, 0) means c=4c = 4.

Step 3: b2=a2c2=3616=20b^2 = a^2 - c^2 = 36 - 16 = 20.

x236+y220=1\frac{x^2}{36} + \frac{y^2}{20} = 1

Example 4: Converting from General Form

Convert 4x2+9y216x+54y+61=04x^2 + 9y^2 - 16x + 54y + 61 = 0 to standard form.

Step 1 — Group and factor:

4(x24x)+9(y2+6y)=614(x^2 - 4x) + 9(y^2 + 6y) = -61

Step 2 — Complete the square:

4(x24x+4)+9(y2+6y+9)=61+16+814(x^2 - 4x + 4) + 9(y^2 + 6y + 9) = -61 + 16 + 81

4(x2)2+9(y+3)2=364(x - 2)^2 + 9(y + 3)^2 = 36

Step 3 — Divide by 36:

(x2)29+(y+3)24=1\frac{(x - 2)^2}{9} + \frac{(y + 3)^2}{4} = 1

Center: (2,3)(2, -3), a2=9a^2 = 9 (under xx), b2=4b^2 = 4 (under yy). Major axis is horizontal. a=3a = 3, b=2b = 2, c=5c = \sqrt{5}.

Graphing an Ellipse

  1. Find the center from the equation
  2. Identify aa and bb — the larger denominator is a2a^2
  3. Plot the center, then move aa units along the major axis in both directions (vertices) and bb units along the minor axis (co-vertices)
  4. Sketch the oval through these four points
  5. Mark the foci at distance cc from center along the major axis

The U.S. Capitol’s National Statuary Hall has an elliptical ceiling. Sound from one focus travels to the other focus with minimal loss — a person whispering at one focus can be heard clearly at the other, even from 40+ feet away.

If the hall’s ceiling is modeled by:

x2625+y2400=1\frac{x^2}{625} + \frac{y^2}{400} = 1

then a=25a = 25 feet, b=20b = 20 feet, and c=625400=225=15c = \sqrt{625 - 400} = \sqrt{225} = 15 feet. The two “sweet spots” for whispering are 15 feet from the center along the major axis.

Common Mistakes

  1. Confusing which denominator is a2a^2. Always: a2a^2 is the larger denominator, regardless of whether it is under xx or yy.
  2. Using c2=a2+b2c^2 = a^2 + b^2. That is the hyperbola formula. For ellipses: c2=a2b2c^2 = a^2 - b^2.
  3. Forgetting to divide when converting from general form. After completing the square, both sides must be divided so the right side equals 1.
  4. Placing foci along the minor axis. Foci are always on the major axis.

Practice Problems

Problem 1: Find the center, vertices, and foci of (x+2)249+(y1)225=1\frac{(x+2)^2}{49} + \frac{(y-1)^2}{25} = 1.

Center: (2,1)(-2, 1). a2=49a^2 = 49 (under xx), b2=25b^2 = 25. Horizontal major axis.

a=7a = 7, b=5b = 5, c=4925=24=26c = \sqrt{49 - 25} = \sqrt{24} = 2\sqrt{6}.

Vertices: (2±7,1)=(9,1)(-2 \pm 7, 1) = (-9, 1) and (5,1)(5, 1)

Foci: (2±26,1)(6.90,1)(-2 \pm 2\sqrt{6}, 1) \approx (-6.90, 1) and (2.90,1)(2.90, 1)

Problem 2: Write the equation of an ellipse with foci at (0,±3)(0, \pm 3) and vertices at (0,±5)(0, \pm 5).

Vertical major axis (foci on yy-axis). a=5a = 5, c=3c = 3.

b2=a2c2=259=16b^2 = a^2 - c^2 = 25 - 9 = 16

x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1

Problem 3: Convert 9x2+4y236x+8y+4=09x^2 + 4y^2 - 36x + 8y + 4 = 0 to standard form.

Group: 9(x24x)+4(y2+2y)=49(x^2 - 4x) + 4(y^2 + 2y) = -4

Complete the square: 9(x24x+4)+4(y2+2y+1)=4+36+49(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = -4 + 36 + 4

9(x2)2+4(y+1)2=369(x - 2)^2 + 4(y + 1)^2 = 36

Divide by 36: (x2)24+(y+1)29=1\frac{(x-2)^2}{4} + \frac{(y+1)^2}{9} = 1

Center: (2,1)(2, -1). Vertical major axis (a2=9a^2 = 9). a=3a = 3, b=2b = 2, c=5c = \sqrt{5}.

Problem 4: An ellipse has eccentricity e=0.6e = 0.6 and semi-major axis a=10a = 10. Find bb and cc.

c=ea=0.6×10=6c = ea = 0.6 \times 10 = 6

b2=a2c2=10036=64b^2 = a^2 - c^2 = 100 - 36 = 64, so b=8b = 8.

Problem 5: The orbit of a planet is an ellipse with the sun at one focus. If a=150a = 150 million km and c=2.5c = 2.5 million km, find the eccentricity and farthest distance from the sun.

e=ca=2.51500.0167e = \frac{c}{a} = \frac{2.5}{150} \approx 0.0167

Farthest distance (aphelion) = a+c=150+2.5=152.5a + c = 150 + 2.5 = 152.5 million km.

Key Takeaways

  • An ellipse is the set of points where the sum of distances to two foci equals 2a2a
  • Standard form: (xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 (horizontal) or denominators swapped (vertical) — the larger denominator is always a2a^2
  • The relationship c2=a2b2c^2 = a^2 - b^2 connects semi-major axis, semi-minor axis, and focal distance
  • Eccentricity e=c/ae = c/a measures elongation: 00 is a circle, close to 11 is very elongated
  • A circle is a special ellipse with a=ba = b (and c=0c = 0)

Return to College Algebra for more topics in this section.

Last updated: March 29, 2026