Ellipses
An ellipse is a stretched circle. The planets move around the Sun on ellipses. You can draw one with two pins, a loop of string, and a pencil: the pins are the foci, and the string keeps the total distance to them fixed.
The definition
Take two points ${F_1}$ and ${F_2}$, called the foci (singular: focus). An ellipse is the set of all points ${P}$ for which the sum of the distances to the foci is a constant, ${2a}$:
${PF_1+PF_2=2a}$
The longest line across the ellipse, through both foci, is the major axis. Its ends are the vertices. The midpoint between the foci is the center. The shorter line across, through the center, is the minor axis.
Standard equations
Put the center at the origin and the foci at ${(\pm c,0)}$. Writing ${PF_1+PF_2=2a}$ with the distance formula, and simplifying (it takes two squarings), gives the equation below, with ${b^2=a^2-c^2}$.
${\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$ with ${a>b}$: wide, foci ${(\pm c,0)}$
${\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1}$ with ${a>b}$: tall, foci ${(0,\pm c)}$
In both: ${c^2=a^2-b^2}$
The larger denominator, ${a^2}$, is under the variable of the major axis. The vertices are ${a}$ units from the center along the major axis, and the ends of the minor axis are ${b}$ units from the center.
Example 1: Find the vertices and foci of ${\dfrac{x^2}{25}+\dfrac{y^2}{9}=1}$.
Solution:
The larger denominator, ${25}$, is under ${x^2}$, so the major axis is horizontal. Then ${a=5}$ and ${b=3}$, and
${c^2=25-9=16}$, so ${c=4}$
The vertices are ${(\pm 5,0)}$ and the foci are ${(\pm 4,0)}$. The ellipse also passes through ${(0,\pm 3)}$. It is the ellipse in the first figure.
Example 2: Find the vertices and foci of ${4x^2+y^2=16}$, and sketch it.
Solution:
Divide by ${16}$ to get ${1}$ on the right:
${\dfrac{x^2}{4}+\dfrac{y^2}{16}=1}$
The larger denominator is under ${y^2}$, so the ellipse is tall: ${a=4}$, ${b=2}$. Then ${c^2=16-4=12}$, so ${c=2\sqrt{3}}$. The vertices are ${(0,\pm 4)}$, and the foci are ${(0,\pm 2\sqrt{3})}$.
Example 3: Find the equation of the ellipse with vertices ${(\pm 6,0)}$ and foci ${(\pm 4,0)}$.
Solution:
The major axis is horizontal, with ${a=6}$ and ${c=4}$. So ${b^2=a^2-c^2=36-16=20}$, and
${\dfrac{x^2}{36}+\dfrac{y^2}{20}=1}$
Eccentricity
How stretched an ellipse is depends on how far apart the foci are. The eccentricity measures this:
${e=\dfrac{c}{a}}$, with ${0<e<1}$
When ${e}$ is near ${0}$, the foci are near the center, and the ellipse is almost a circle. When ${e}$ is near ${1}$, it is long and thin.
The Earth's orbit has ${e\approx 0.017}$: nearly a circle. Halley's comet has ${e\approx 0.97}$: a very long, thin ellipse.
Shifted ellipses
With center ${(h,k)}$, replace ${x}$ by ${x-h}$ and ${y}$ by ${y-k}$. To find the center from an expanded equation, complete the square in ${x}$ and in ${y}$.
Example 4: Find the center and sketch ${9x^2+4y^2-18x+16y-11=0}$.
Solution:
Group the terms, and factor out the coefficients of ${x^2}$ and ${y^2}$:
${9(x^2-2x)+4(y^2+4y)=11}$
Complete each square. Adding ${1}$ inside the first group adds ${9\cdot 1=9}$, and adding ${4}$ inside the second adds ${4\cdot 4=16}$:
$\begin{align*}9(x-1)^2+4(y+2)^2&=11+9+16\\9(x-1)^2+4(y+2)^2&=36\end{align*}$
Divide by ${36}$:
${\dfrac{(x-1)^2}{4}+\dfrac{(y+2)^2}{9}=1}$
The center is ${(1,-2)}$. The ellipse reaches ${3}$ units up and down and ${2}$ units left and right. Here ${c=\sqrt{9-4}=\sqrt{5}}$, so the foci are ${(1,-2\pm\sqrt{5})}$.
Summary
- An ellipse is the set of points whose distances to two foci add up to ${2a}$.
- ${\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$ (${a>b}$): vertices ${(\pm a,0)}$, foci ${(\pm c,0)}$, with ${c^2=a^2-b^2}$. If the larger denominator is under ${y^2}$, the ellipse is tall.
- Eccentricity ${e=\dfrac{c}{a}}$ is between ${0}$ (a circle) and ${1}$ (very thin).
- For a shifted ellipse, complete the square to find the center ${(h,k)}$.