Hyperbolas
A hyperbola looks like two parabolas facing away from each other, but it is a different curve. It appears in the path of a spacecraft swinging past a planet, in the shape of cooling towers, and in navigation systems that locate a ship by timing radio signals.
The definition
Take two foci ${F_1}$ and ${F_2}$. A hyperbola is the set of all points ${P}$ for which the difference of the distances to the foci is a constant, ${2a}$:
${|PF_1-PF_2|=2a}$
A hyperbola has two separate pieces, called branches. The points where it crosses the line through the foci are the vertices, and the midpoint of the foci is the center.
Standard equations
With the center at the origin and foci at ${(\pm c,0)}$, the definition leads to the equation below, now with ${b^2=c^2-a^2}$. (Compare the ellipse, where ${b^2=a^2-c^2}$.)
${\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}$: opens left and right, vertices ${(\pm a,0)}$, foci ${(\pm c,0)}$
${\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1}$: opens up and down, vertices ${(0,\pm a)}$, foci ${(0,\pm c)}$
In both: ${c^2=a^2+b^2}$
The hyperbola opens along the axis of the positive term. Unlike the ellipse, ${a}$ need not be larger than ${b}$.
Asymptotes
Far from the center, the branches get closer and closer to two straight lines, the asymptotes. For ${\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}$ they are
${y=\pm\dfrac{b}{a}x}$
An easy way to draw them: sketch the box with sides through ${x=\pm a}$ and ${y=\pm b}$. The asymptotes are the lines through its corners. For the up-and-down hyperbola ${\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1}$, they are ${y=\pm\dfrac{a}{b}x}$.
Example 1: Find the vertices, foci, and asymptotes of ${\dfrac{x^2}{16}-\dfrac{y^2}{9}=1}$.
Solution:
The positive term has ${x^2}$, so the hyperbola opens left and right. Here ${a=4}$ and ${b=3}$, so
${c^2=16+9=25}$, so ${c=5}$
The vertices are ${(\pm 4,0)}$, the foci are ${(\pm 5,0)}$, and the asymptotes are ${y=\pm\dfrac{3}{4}x}$. This is the hyperbola in the first figure.
Example 2: Find the vertices, foci, and asymptotes of ${\dfrac{y^2}{9}-\dfrac{x^2}{16}=1}$.
Solution:
The positive term has ${y^2}$, so it opens up and down, with ${a=3}$ and ${b=4}$. Again ${c=\sqrt{9+16}=5}$. The vertices are ${(0,\pm 3)}$, the foci are ${(0,\pm 5)}$, and the asymptotes are ${y=\pm\dfrac{3}{4}x}$.
Example 3: Find the equation of the hyperbola with vertices ${(0,\pm 2)}$ and foci ${(0,\pm 3)}$.
Solution:
It opens up and down, with ${a=2}$ and ${c=3}$. So ${b^2=c^2-a^2=9-4=5}$, and
${\dfrac{y^2}{4}-\dfrac{x^2}{5}=1}$
Watch out: For an ellipse, ${c^2=a^2-b^2}$. For a hyperbola, ${c^2=a^2+b^2}$. The foci of a hyperbola are always farther from the center than its vertices.
Shifted hyperbolas
As before, a center at ${(h,k)}$ means replacing ${x}$ by ${x-h}$ and ${y}$ by ${y-k}$. Complete the square to find it.
Example 4: Find the center, vertices, and asymptotes of ${4x^2-y^2-8x-4y-4=0}$.
Solution:
Group and factor:
${4(x^2-2x)-(y^2+4y)=4}$
Complete the squares. Adding ${1}$ inside the first group adds ${4}$; adding ${4}$ inside the second group subtracts ${4}$ (because of the minus sign in front):
$\begin{align*}4(x-1)^2-(y+2)^2&=4+4-4\\4(x-1)^2-(y+2)^2&=4\end{align*}$
Divide by ${4}$:
${(x-1)^2-\dfrac{(y+2)^2}{4}=1}$
The center is ${(1,-2)}$, with ${a=1}$ and ${b=2}$. It opens left and right, so the vertices are ${(0,-2)}$ and ${(2,-2)}$. The asymptotes pass through the center with slopes ${\pm\dfrac{b}{a}=\pm 2}$:
${y+2=\pm 2(x-1)}$
Summary
- A hyperbola is the set of points whose distances to two foci differ by ${2a}$.
- ${\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}$: vertices ${(\pm a,0)}$, foci ${(\pm c,0)}$ with ${c^2=a^2+b^2}$, asymptotes ${y=\pm\dfrac{b}{a}x}$.
- It opens along the axis of the positive term.
- Draw the box through ${x=\pm a}$, ${y=\pm b}$; the asymptotes go through its corners.