Results 1 to 3 of 3

Math Help - Help with hyperbola's equations.

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    4

    Help with hyperbola's equations.

    I have two question on an assignment of mine that I just can't seem to figure out, can you help me?

    Write an equation of a hyperbola with vertices (1, 4) and (-5, 4), and foci (3, 4) and (-7, 4).

    and

    Write an equation of a hyperbola with vertices (8, 2) and (-4, 2), and foci (12,2) and (-8, 2).


    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    so you know that the vertex are (1, 4),(-5, 4) so you find the center(x,y). with midpoint formula.

    <br />
(x,y) = \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}<br />

    in your case center is (-2,4) so lets call this distance from center to vertex 'A'.
    -5-(-2)=|-3| , 1-(-2)=|3|, A=3

    now you can use hyperbola formula to derive standard form.
    you know that A=3 and you have foci F at (3, 4),(-7, 4).

    <br />
| \sqrt{(X_1-FX_1)^2+(Y_1-FY_1)^2} - \sqrt{(X_2-FX_2)^2+(Y_2-FY_2)^2} | = 2A<br />
    Last edited by skoker; February 13th 2011 at 05:08 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,690
    Thanks
    617
    Hello, maramfaith24!

    Write an equation of a hyperbola with vertices (1, 4) and (-5, 4)
    and foci (3, 4) and (-7, 4).

    You're expected to know the general forms of the hyperbola.

    . . \displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} \:=\:1 \qquad \text{"horizontal" hyperbola: }\supset \subset

    . . \displaystyle \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} \;=\;1 \qquad\text{"vertical" hyperbola: }\begin{array}{c}\cup \\ [-2mm]\cap \end{array}

    . . where (h,k) is the center of the hyperbola
    . . and the focal distance \,c is given by: . a^2+b^2 \:=\:c^2 .[1]



    The vertices are (1,4) and (\text{-}5,4), on a horizontal line.
    . . We have a "horizontal" hyperbola.

    The center is the midpoint of the vertices: . (h,k) \,=\,(\text{-}2,4)

    The distance from the center (\text{-}2,4) to a vertex (1,4) is: . 1 - (\text{-}2)
    . . Hence: a = 3

    The distance from the center (\text{-}2,4) to a focus (3,4) is: . 3- (\text{-}2)
    . . Hence: c = 5

    Substitute into [1]: . 3^2 + b^2 \:=\:5^2 \quad\Rightarrow\quad b^2 \,=\,16



    We know the center of the hyperbola: . (h,k) \:=\:(\text{-}2,4)
    . . We know that: . a^2 = 9,\;b^2 = 16
    . . And we know that the hyperbola is "horizontal".

    \displaystyle \text{The equation is: }\;\;\frac{(x+2)^2}{9} - \frac{(y-4)^2}{16} \;=\;1

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 15th 2010, 11:15 PM
  2. hyperbola help!
    Posted in the Geometry Forum
    Replies: 4
    Last Post: May 5th 2010, 10:29 PM
  3. Hyperbola and Elipse Equations
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 12th 2009, 01:23 PM
  4. need help with hyperbola equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 22nd 2009, 02:22 AM
  5. Hyperbola
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 20th 2008, 06:43 AM

Search Tags


/mathhelpforum @mathhelpforum