Graphing Hyperbolas

Key Questions

  • What information do you need to graph hyperbolas?

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

    #(x-x_c)^2/a^2-(y-y_c)^2/b^2=+-1#,

    we can graph the hyperbolas in this way:

    • find the center #C(x_c,y_c)#;

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

    • 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.

    Massimiliano · 1 · Mar 13 2015
  • How do I find an equation for a hyperbola, given its graph?

    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#

    Yahia M. · 1 · Jun 16 2018

Questions

Geometry of a Hyperbola