Results 1 to 4 of 4

Math Help - Parabolas- Need help

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    44

    Exclamation Parabolas- Need help

    Can you please help me with the following questions.

    Graph the parabola with the given equation.

    (y-2)^2=6(x-1)
    I know the vertex would be (1,2)

    (x+1)^2=6(y-1)
    The vetex would be (-1,1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by soly_sol View Post
    Can you please help me with the following questions.

    Graph the parabola with the given equation.

    (y-2)^2=6(x-1)
    I know the vertex would be (1,2)
    yes, and where should be the focus?
    note that the distance of the focus from the vertex of the parabola (y-k)^2 = 4p(x-h) is p. what is your p in your equation?
    also note that the distance of the endpoints of the latus rectum is 4p, or teh distance of one endpoint from the focus is 2p.. (by the way, remember that the latus rectum is the segment perpendicular to the axis of symmetry of the parabola passing through the focus)

    with this information, i hope you could sketch the graphs..


    Quote Originally Posted by soly_sol View Post
    (x+1)^2=6(y-1)
    The vetex would be (-1,1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2007
    Posts
    44
    Graph the parabola with the given equation.

    (y-2)^2=6(x-1)
    I know the vertex would be (1,2)
    Here else is what I got:
    I got p=24
    foci: (25,2)
    directrix: x=-23
    axis:y=2

    (x+1)^2=6(y-1)
    The vetex would be (-1,1)
    foci: (-1, 25)
    directrix: y=-23
    axis: -1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by soly_sol View Post
    Graph the parabola with the given equation.

    (y-2)^2=6(x-1)
    I know the vertex would be (1,2)
    Here else is what I got:
    I got p=24
    foci: (25,2)
    directrix: x=-23
    axis:y=2
    4p = 6 \implies p = \frac{3}{2}
    from here, do the rests...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parabolas please help!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 4th 2010, 10:31 AM
  2. Parabolas
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: January 8th 2009, 07:47 AM
  3. I Need Help With Parabolas!!!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 26th 2008, 03:42 PM
  4. Parabolas
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 5th 2008, 12:25 PM
  5. parabolas
    Posted in the Pre-Calculus Forum
    Replies: 14
    Last Post: April 1st 2008, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum