Ellipses in Standard Form

Learn to write, graph, and analyze ellipses using the standard form equation.

Advanced25 minLesson
Printable worksheet

Definition

An ellipse is the set of all points (x,y)(x, y) such that the sum of the distances from two fixed points (called foci) is constant.

Standard Form Equations

Horizontal ellipse (wider than tall):
(x−h)2a2+(y−k)2b2=1where a>b\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad \text{where } a > b
Vertical ellipse (taller than wide):
(x−h)2b2+(y−k)2a2=1where a>b\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \quad \text{where } a > b

Key Components

  • Center: (h,k)(h, k)
  • Semi-major axis: aa (always the larger value)
  • Semi-minor axis: bb (always the smaller value)
  • Relationship: c2=a2−b2c^2 = a^2 - b^2 where cc is the distance from center to each focus

Try it now

What is the center of the ellipse (x−3)216+(y+2)29=1\frac{(x-3)^2}{16} + \frac{(y+2)^2}{9} = 1?

Worked Examples

Find the center, vertices, co-vertices, and foci of the ellipse: (x−2)225+(y+3)29=1\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1
1

Identify the center (h,k)(h, k)

From (x−2)2(x-2)^2 and (y+3)2(y+3)^2, we have h=2h = 2 and k=−3k = -3 → Center: (2,−3)(2, -3)
2

Find a2a^2 and b2b^2

a2=25a^2 = 25 (larger), b2=9b^2 = 9 (smaller) → a=5a = 5, b=3b = 3
3

Determine orientation

Since a2=25a^2 = 25 is under (x−h)2(x-h)^2, the major axis is horizontal → Horizontal ellipse
4

Find vertices (on major axis)

(h±a,k)=(2±5,−3)(h \pm a, k) = (2 \pm 5, -3) → Vertices: (−3,−3)(-3, -3) and (7,−3)(7, -3)
5

Find co-vertices (on minor axis)

(h,k±b)=(2,−3±3)(h, k \pm b) = (2, -3 \pm 3) → Co-vertices: (2,−6)(2, -6) and (2,0)(2, 0)
6

Calculate cc for foci

c2=a2−b2=25−9=16c^2 = a^2 - b^2 = 25 - 9 = 16, so c=4c = 4
7

Find foci (on major axis)

(h±c,k)=(2±4,−3)(h \pm c, k) = (2 \pm 4, -3) → Foci: (−2,−3)(-2, -3) and (6,−3)(6, -3)

Common Mistakes

Confusing which denominator gives a2a^2 vs b2b^2

Why it's wrong: Students assume a2a^2 is always under x2x^2, but a>ba > b by definition, so a2a^2 is the larger denominator.

Correct: Always identify: a2a^2 = larger denominator, b2b^2 = smaller denominator. Then determine orientation based on which variable has a2a^2.

Using c2=a2+b2c^2 = a^2 + b^2 (hyperbola formula) instead of c2=a2−b2c^2 = a^2 - b^2

Why it's wrong: Hyperbolas use addition, ellipses use subtraction. This is a critical difference.

Correct: For ellipses: c2=a2−b2c^2 = a^2 - b^2. Remember: foci are inside the ellipse, so c<ac < a.

Forgetting to change signs when reading center from equation

Why it's wrong: The equation has (x−h)2(x - h)^2, so (x+3)2(x + 3)^2 means h=−3h = -3, not h=3h = 3.

Correct: (x−h)2(x - h)^2 with (x+3)2(x + 3)^2 means −h=3-h = 3, so h=−3h = -3.

Placing foci on the minor axis

Why it's wrong: Students may forget that foci are always on the major axis (the longer one).

Correct: Foci are always on the major axis, at distance cc from the center.

Watch a video explanation

The same topic explained by another teacher, if a video helps you more.

Ellipses Explained: Vertex, Foci, Eccentricity, and Equations

by Understand The Math

Watch on YouTube

Interactive Visual

Function graph

-6-4-20246810-10-50xyCF₁F₂
  • c = √|a² − b²| = 4
(x − h)²/a² + (y − k)²/b² = 1
  • Center: (2, -3)
  • Foci: (-2, -3), (6, -3)
  • Vertices: (-3, -3), (7, -3)

Interactive Sandbox

Interactive Grapher

-10-50510-10-50510xy

Tap or hover over the graph to read off values.

Try these examples:

y = -3 + 3*sqrt(1 - (x - 2)^2/25)

y = -3 - 3*sqrt(1 - (x - 2)^2/25)

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Practice Problems

17 problems
Problem 1 of 17
Easy
What is the center of the ellipse (x−3)216+(y+2)29=1\frac{(x-3)^2}{16} + \frac{(y+2)^2}{9} = 1?

Why It Matters

Ellipses appear throughout science and engineering:
  • Astronomy: Planets orbit the Sun in elliptical paths (Kepler's First Law)
  • Architecture: Whispering galleries use elliptical ceilings for acoustic effects
  • Medicine: Lithotripsy machines use elliptical reflectors to focus sound waves on kidney stones
  • Optics: Elliptical mirrors reflect all light from one focus to the other
Understanding the standard form helps us analyze and design these systems mathematically.

Real World Applications

Planetary Orbits

All planets orbit the Sun in elliptical paths with the Sun at one focus.

Example:

Earth's orbit has a≈149.6a \approx 149.6 million km and eccentricity e≈0.017e \approx 0.017, making it nearly circular.

1Try It Yourself

A comet orbits the Sun in an ellipse with semi-major axis a=18a = 18 AU and c=17c = 17 AU (distance from center to Sun).

Find the semi-minor axis bb.

Step 1: Write the mathematical expression

Use c2=a2−b2c^2 = a^2 - b^2:

Whispering Galleries

In an elliptical room, sound from one focus reflects to the other focus, allowing whispers to travel across the room.

Example:

The Capitol Building in Washington D.C. has an elliptical room where this effect occurs.

2Try It Yourself

An elliptical hall is 30 meters long and 20 meters wide.

How far apart are the two focal points (whispering spots)?

Step 1: Write the mathematical expression

Find 2c2c where a=15a = 15 and b=10b = 10:

Key Takeaways

  • 1An ellipse in standard form is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 (horizontal) or with a2a^2 and b2b^2 swapped (vertical)
  • 2The center is at (h,k)(h, k), and a>ba > b always (larger denominator determines orientation)
  • 3Vertices are on the major axis at distance aa from center; co-vertices are on minor axis at distance bb
  • 4Foci are on the major axis at distance cc from center, where c2=a2−b2c^2 = a^2 - b^2
  • 5To convert general form to standard form, complete the square for both variables

Frequently Asked Questions

Look at which variable has the larger denominator. If a2a^2 (larger) is under (x−h)2(x-h)^2, the major axis is horizontal. If a2a^2 is under (y−k)2(y-k)^2, the major axis is vertical.
Look at which variable has the larger denominator. If a2a^2 (larger) is under (x−h)2(x-h)^2, the major axis is horizontal. If a2a^2 is under (y−k)2(y-k)^2, the major axis is vertical.
aa is the distance from center to vertex (on the major axis). cc is the distance from center to focus. Since foci are inside the ellipse, c<ac < a.
When a=ba = b, the ellipse becomes a circle! The equation becomes (x−h)2+(y−k)2=a2(x-h)^2 + (y-k)^2 = a^2, and there are no distinct foci (c=0c = 0).

Glossary

Ellipse
The set of all points where the sum of distances to two foci is constant
Semi-major axis
Half the length of the longest diameter, denoted aa
Semi-minor axis
Half the length of the shortest diameter, denoted bb
Focus (pl. foci)
One of two fixed points inside the ellipse; c2=a2−b2c^2 = a^2 - b^2
Vertex
An endpoint of the major axis
Co-vertex
An endpoint of the minor axis
Eccentricity
The ratio e=c/ae = c/a measuring how elongated the ellipse is (0 = circle, close to 1 = very elongated)

Formula Card

Standard Form (horizontal)

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

$a > b$, major axis horizontal

Standard Form (vertical)

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

$a > b$, major axis vertical

Relationship

c2=a2−b2c^2 = a^2 - b^2

$c$ = distance from center to focus

Eccentricity

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

$0 < e < 1$ for all ellipses

Today's challengeOne problem, three tries

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