# More on hyperbolas

A
hyperbola
is the set of all points
$P$
in the plane such that the difference between the distances from
$P$
to two fixed points is a given constant. Each of the fixed points is a
focus
. (The plural is foci.) If
$P$
is a point on the hyperbola and the foci are
${F}_{1}$
and
${F}_{2}$
then
$\stackrel{\xaf}{P{F}_{1}}$
and
$\stackrel{\xaf}{P{F}_{2}}$
are the
**
focal radii
**
. As you can see, the graph of the hyperbola has two disconnected branches that look similar to parabolas. Each branch approaches diagonal
asymptotes
.

The
**
center
**
of a hyperbola is the midpoint of the line segment joining its foci. The
**
transverse axis
**
is the line segment that contains the center of the hyperbola and whose endpoints are the two vertices of the hyperbola.

A central hyperbola, one with its center on the origin and its foci on either the $x$ -axis or $y$ -axis has one to the two formulas below. Note that ${a}^{2}$ is the denominator of the positive term in each case.

The hyperbola with its center at $\left(0,0\right)$ , foci $\left(c,0\right)$ and $\left(-c,0\right)$ and distance between the vertices $2a$ has the equation

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ where ${b}^{2}={c}^{2}-{a}^{2}$ .

The equations of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

Note that the hyperbola will have either $x$ - or $y$ -intercepts but never both.

The hyperbola with its center at $\left(0,0\right)$ , foci $\left(0,c\right)$ and $\left(0,-c\right)$ , then the equation is $\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1$ with ${b}^{2}={c}^{2}-{a}^{2}$

The equations of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$ .

**
Example 1:
**

Find an equation of the hyperbola having foci $\left(3,0\right)$ and $\left(-3,0\right)$ and the distance between the vertices equal to $4$ if its center is at the origin.

The distance from each focus to the center is $3$ , so $c=3$ .

The distance between the vertices is $2a$ . So, $2a=4$ and $a=2$ .

${b}^{2}={c}^{2}-{a}^{2}$ , so ${b}^{2}={3}^{2}-{2}^{2}=9-4=5$

Since the foci are on the $x$ -axis, the equation is $\frac{{x}^{2}}{4}-\frac{{y}^{2}}{5}=1$ .

**
Example 2:
**

For the above example, find the equations of its asymptotes and graph the hyperbola.

The equations of the asymptotes are $y=\frac{\sqrt{5}}{2}x$ and $y=-\frac{\sqrt{5}}{2}x$ .

The $x$ -intercepts are $\left(2,0\right)$ and $\left(-2,0\right)$ . There are no $y$ -intercepts.

The graph of a hyperbola can be translated so that its center is at the point $\left(h,k\right)$ . This means that the graph has been translated $h$ units on the horizontal axis and $k$ units on the vertical axis.

*
Horizontal Major Axis Vertical Major Axis
*

$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ $\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$

Foci at $\left(h-c,k\right)$ and $\left(h+c,k\right)$ Foci at $\left(h,k-c\right)$ and $\left(h,k+c\right)$

**
Example 3:
**

Find an equation of the hyperbola with foci $\left(-3,-2\right)$ and $\left(-3,8\right)$ and the distance between the vertices $8$ .

The distance between the vertices, $2a$ , is $8$ . So, $2a=8$ and $a=4$ .

The center is halfway between the foci at $\left(-3,3\right)$ .

The distance from the center to each focus is $5$ so $c=5$ .

${b}^{2}={c}^{2}-{a}^{2}={5}^{2}-{4}^{2}=25-16=9$

Therefore, the equation of the hyperbola is $\frac{{\left(y-3\right)}^{2}}{16}-\frac{{\left(x+3\right)}^{2}}{9}=1$

A hyperbola can also be defined as a conic section obtained by the intersection of a double cone with a plane that is intersects both pieces of the cone without intersecting the axis.