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Pre-Calculus Conic sections

Ellipses (standard form, foci, center at (h, k))

20 practice questions 0 video lessons Theory + worked examples

Ellipses

Texas Precalculus (TEKS) • Standard P.3(H) • Conic Sections

Ellipses is a topic in Conic Sections in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.3(H), which requires students to write the equation of an ellipse with center (h, k).

An ellipse is the set of points whose distances to two foci add to a constant, with standard form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

Texas Precalculus (TEKS) › Conic Sections › Ellipses  —  Standard P.3(H)

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Theory

An ellipse is the set of points whose distances to two fixed foci add to a constant. Centered at the origin its standard form is

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\qquad a>b.\]

The major axis (length \(2a\)) lies along the variable with the larger denominator; the minor axis has length \(2b\). The foci sit a distance \(c\) from the center, where

\[c^2=a^2-b^2.\]
The larger denominator points to the major axis. If it is under \(y^2\), the ellipse is taller than it is wide.
Ellipse with two foci An ellipse is the set of points whose distances to two foci sum to a constant. x y focus focus
An ellipse: the sum of distances to the two foci is constant.
Standard form (center origin) Standard form (center origin) Standard form (center origin) + = 1 a > b: major axis horizontal c² = a² − b² (foci at ±c)
Standard form and the foci.

Standard form and the focal distance:

\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\qquad c^2=a^2-b^2\]
x squared over a squared plus y squared over b squared equals 1; c squared equals a squared minus b squared
Center \((h,k)\): replace \(x,y\) with \(x-h,\ y-k\).

How to analyze an ellipse

  1. Identify \(a^2\) and \(b^2\) (\(a^2\) is the larger).
  2. Major axis lies along the larger-denominator variable, length \(2a\).
  3. Foci: \(c=\sqrt{a^2-b^2}\) from the center, along the major axis.
  4. Center from the \((x-h),(y-k)\) shifts.
Example 1 — Axes and foci
For \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\), find the axes lengths and foci.
Solution

\(a^2=25,\ b^2=9\), so \(a=5,\ b=3\); \(c^2=25-9=16\).

\(\text{major axis}\)\(=\)\(2a=10\)
\(\text{minor axis}\)\(=\)\(2b=6\)
\(\text{foci}\)\(=\)\((\pm 4,0)\)
major axis 10, minor axis 6, foci at plus or minus 4
Example 2 — Vertical major axis
Describe \(\dfrac{x^2}{9}+\dfrac{y^2}{25}=1\).
Solution

Here the larger denominator is under \(y^2\), so the major axis is vertical.

\(a=5\ (\text{on }y),\ b=3\)
\(\text{foci}\)\(=\)\((0,\pm 4)\)
major axis vertical, foci at 0 comma plus or minus 4
Example 3 — Write an equation
Write the ellipse centered at the origin with \(a=6\) horizontal and \(b=4\).
Solution

Place \(a^2\) under \(x^2\).

\[\dfrac{x^2}{36}+\dfrac{y^2}{16}=1\]
equation is x squared over 36 plus y squared over 16 equals 1
Example 4 — Center at (h, k)
Give the center of \(\dfrac{(x-2)^2}{9}+\dfrac{(y+1)^2}{4}=1\).
Solution

Read the shifts directly.

\(\text{center}\)\(=\)\((2,-1)\)
center is 2 comma negative 1

Common pitfalls

\(a^2\) is the larger denominator, and it decides the major axis's direction.
Foci use \(c^2=a^2-b^2\) (subtract), unlike the hyperbola.
The foci lie on the major axis, not the minor one.

Frequently asked questions

What is an ellipse?

The set of points whose distances to two foci sum to a constant; its standard form is \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

How do you find the foci of an ellipse?

Compute \(c=\sqrt{a^2-b^2}\); the foci lie a distance \(c\) from the center along the major axis.

How do you tell which axis is major?

The larger denominator marks the major axis. If it is under \(y^2\), the ellipse is taller; under \(x^2\), it is wider.

What is the equation of a shifted ellipse?

\(\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\), centered at \((h,k)\).