Results 1 to 4 of 4

Math Help - Hey gues find equation of parabola with given properties. Also a hyperbola question.

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    1

    Hey gues find equation of parabola with given properties. Also a hyperbola question.

    Hey guys I have a question I've been stuck on. The problem says

    Find the equation of the parabola with the given properties.
    Vertex(1,3); vertical axis of symmetry; containing the point(-1,19)

    I can't find this in my book anywhere.

    I also have on that says Find the center, foci, and asymptotes of the hyperbola.

    7x^2 - y^2 - 42x + 8y - 65 = 0

    I think I have the completing the square down. I'm not sure though
    I get stuck at this part

    7(x^2 - 6x) - 1(y^2 - 8y) = -65

    I'm stuck what to do with the completing the square after that. After that I'm not sure on how to find the center, asymptotes and foci.

    And I hope I'm not asking to much. But I have this question to. I'm not sure where to start.

    Find the equation of the hyperbola satisfying the given conditions.
    Vertices: (-6,4) and (8,4); asymptotes y = 2x + 8 and y = -2x + 0


    Thanks for any help, steps, or answers guys. I appreciate a lot.
    Last edited by dham; April 29th 2009 at 07:43 PM. Reason: Adding a question
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by dham View Post
    Hey guys I have a question I've been stuck on. The problem says

    Find the equation of the parabola with the given properties.
    Vertex(1,3); vertical axis of symmetry; containing the point(-1,19)
    I would use the equation for turning point

    y=a(x-h)^2+k

    Where a is the dialation (determines the steepness of the function), h is the horizontal translation and k is the vertical translation.

    If the vertex is at (1,3) then we can say

    y=a(x-1)^2+3 now substitute the point (-1,19) into this equation to find a.

    19=a(-1-1)^2+3

    19=a(-2)^2+3

    19=4a+3

    16=4a

     a=4


    therefore the equation for this parabola is,

    y=4(x-1)^2+3
    Last edited by pickslides; April 29th 2009 at 09:47 PM. Reason: Typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member pflo's Avatar
    Joined
    Apr 2009
    From
    Albuquerque, NM
    Posts
    155
    Thanks
    7

    Completing the Square & Hyperbolas

    Quote Originally Posted by dham View Post

    I think I have the completing the square down. I'm not sure though.
    I get stuck at this part: 7(x^2 - 6x) - 1(y^2 - 8y) = -65

    I'm stuck what to do with the completing the square after that.
    After that I'm not sure on how to find the center, asymptotes and foci.
    You would have 7[(x-3)^2-9]-1[(y-4)^2-16]=-65 after completing the square.

    Rearrange to put this into the standard form for a hyperbola: \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 (standard form when the 'y-squared' term is negative, which results in a horizontal transverse axis - the equations below change slightly if the hyperbola has a vertical transverse axis).

    The center of the hyperbola is at (h,k) and both asymptotes go through this point. The slopes of the asymptotes are {\pm}\frac{b}{a}
    The vertices are at ({\pm}a+h,k))
    The foci are at ({\pm}c+h,k) where c=\sqrt{a^2+b^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by dham View Post
    I can't find this in my book anywhere.
    Ouch!

    To replace the missing lessons, try studying from some online resources. Click for explanations of parabolas and hyperbolas. If you're also needing help with the other conics, click for explanations of circles and ellipses.

    Good luck!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  2. hyperbola or parabola?
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: December 6th 2009, 06:56 PM
  3. Parabola, Elipses, Hyperbola help?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 23rd 2009, 01:56 AM
  4. Parabola Ellipse and Hyperbola
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 11th 2009, 03:14 AM
  5. Some Hyperbola, ellipse, parabola help.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 26th 2008, 03:18 AM

Search Tags


/mathhelpforum @mathhelpforum