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

1 Answer
Jun 16, 2018

Answer:

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#