# Graphing Hyperbolas

## Key Questions

• If It's known the equation of the hyperbolas, that is:

${\left(x - {x}_{c}\right)}^{2} / {a}^{2} - {\left(y - {y}_{c}\right)}^{2} / {b}^{2} = \pm 1$,

we can graph the hyperbolas in this way:

• find the center $C \left({x}_{c} , {y}_{c}\right)$;

• make a rectangle with the center in $C$ and with sides $2 a$ and $2 b$;

• draw the lines that pass from the opposite vertices of the rectangle (the asymptotes);

• if the sign of $1$ is $+$, than the two branches are left and right of the rectangule and the vertices are in the middle of the vertical sides, if the sign of $1$ is $-$, than the two branches are up and down of the rectangule and the vertices are in the middle of the horizontal sides.

check the explanation below.

#### Explanation:

$A {x}^{2} + B {y}^{2} + C x y + D x + E y + F = 0$

That's the general equation of any conic section including the hyperbola

where you can find the equation of a hyperbola given enough points

where color(green)(" A,B,C,D,E,F are constants
color(red)("where either A or B are negative but never both"

though this is usually too hard to solve this type of equation manually

color(blue)("Special Case:"
the hyperbola is horizontal or vertical

if it's horizontal you could use this

color(green)((x-h)^2/a^2-(y-k)^2/(a^2(e^2-1))=1
color(green)("where a is a constant and "e" is the eccentricity "(h,k)" is the center of the hyperbola"

$\textcolor{b l u e}{\text{Example :}}$

graph{(x-1)^2-(y+2)^2/3=1 [-3.502, 5.262, -3.852, 0.533]}
given one of it's foci points is $\left(3 , - 2\right)$

Using the properties of the hyperbola to determine the constants

• the distance between the two vertices equals $2 a$
$2 a = 2$ $\rightarrow$ $a = 1$
• now to get the center $\left(h , k\right)$

it can simply be done by finding the midpoint of the line segment joining the two vertices

$\left(\frac{0 + 2}{2} , \frac{- 2 - 2}{2}\right) = \left(1 , - 2\right)$

• now finally to find $e$

$a e$ is the distance between the center and the focus

$a e = 3 - 1 = 2$

$e = 2$

color(blue)("Finally by substituting"

you get that the hyperbola's equation is as follows

${\left(x - 1\right)}^{2} / 1 - {\left(y + 2\right)}^{2} / 2 = 1$

color(green)("Additional tricks that could help :)"

distance between the directrix and the center of the hyperbola color(green)(a/e

if it's vertical the equation will change but with the same solving technique and it will be

color(green)((y-k)^2/a^2-(x-h)^2/(a^2(e^2-1))=1