Results 1 to 4 of 4
Like Tree2Thanks
  • 2 Post By MarkFL

Thread: Standard form help

  1. #1
    Junior Member
    Joined
    Jul 2018
    From
    Anchorage
    Posts
    41

    Standard form help

    Need to get into standard form.

    4y^2+x^2+7y+2x-6=-3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,101
    Thanks
    807

    Re: Standard form help

    Try completing the square on x and y...what do you get?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2018
    From
    Anchorage
    Posts
    41

    Re: Standard form help

    All I got so far is (+1)^2 + (4y^2 + 7y + 12 1/4)= 16 1/4. I am stuck there.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,101
    Thanks
    807

    Re: Standard form help

    We are given:

    $\displaystyle 4y^2+x^2+7y+2x-6=-3$

    The first thing I would do is add 6 to both sides:

    $\displaystyle 4y^2+x^2+7y+2x=3$

    Next group by like variables on the left:

    $\displaystyle \left(4y^2+7y\right)+\left(x^2+2x\right)=3$

    We see $\displaystyle y^2$ has a coefficient other than 1, so factor that out:

    $\displaystyle 4\left(y^2+\frac{7}{4}y\right)+\left(x^2+2x\right) =3$

    Now add the square of half the coefficients of the linear terms (on both sides):

    $\displaystyle 4\left(y^2+\frac{7}{4}y+\frac{49}{64}\right)+\left (x^2+2x+1\right)=3+4\cdot\frac{49}{64}+1$

    Rewrite:

    $\displaystyle 4\left(y+\frac{7}{8}\right)^2+(x+1)^2=\frac{113}{1 6}$

    Multiply through by $\displaystyle \frac{16}{113}$

    $\displaystyle \frac{64}{113}\left(y+\frac{7}{8}\right)^2+\frac{1 6}{113}(x+1)^2=1$

    Write in standard form:

    $\displaystyle \frac{(x+1)^2}{ \left(\dfrac{\sqrt{113}}{4}\right)^2}+ \frac{\left(y+\dfrac{7}{8}\right)^2}{ \left(\dfrac{\sqrt{113}}{8}\right)^2}=1$
    Thanks from topsquark and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Jan 2nd 2013, 03:14 PM
  2. Replies: 3
    Last Post: Feb 3rd 2011, 06:41 PM
  3. Replies: 2
    Last Post: Mar 9th 2010, 08:33 PM
  4. Replies: 1
    Last Post: Feb 16th 2010, 07:21 AM
  5. Replies: 14
    Last Post: May 30th 2008, 06:10 AM

/mathhelpforum @mathhelpforum