Question

How do I find an equation for a hyperbola, given its graph?

Answer 1

check the explanation below.

Explanation:

`Ax^2+By^2+Cxy+Dx+Ey+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"`

`color(blue)("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 `(3,-2)`

Using the properties of the hyperbola to determine the constants

• the distance between the two vertices equals `2a`
  `2a=2` `rarr` `a=1`
• now to get the center `(h,k)`

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

`((0+2)/2,(-2-2)/2)=(1,-2)`

• now finally to find `e`

`ae` is the distance between the center and the focus

`ae=3-1=2`

`e=2`

`color(blue)("Finally by substituting"`

you get that the hyperbola's equation is as follows

`(x-1)^2/1-(y+2)^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`