Results 1 to 11 of 11

Math Help - Rectanglar hyperbola question

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Rectanglar hyperbola question

    Hi
    can someone show me how would you find the equation express in the form \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1 and asymptotes of the following equations:

    1) x^2-4y^2-4x-8y-16=0
    i got \frac{(x-2)^2}{4}-(y-1)^2=6


    2) 4x^2-8x-y^2+2y=0
    This is what i have done:
    4(x^2-2x)-(y-2y)=0

    (x-1)^2-\frac{(y-1)^2}{4}=0

    answer is \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1

    P.S
    Last edited by mr fantastic; January 15th 2010 at 03:54 PM. Reason: Fixed the general standard form
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634
    Hello, Paymemoney!

    Your algebra is off . . .


    Express in the form \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1
    and find the asymptotes of the following equations:

    1)\;x^2-4y^2-4x-8y-16\:=\:0

    i got: \frac{(x-2)^2}{4}-(y-1)^2\:=\:6 . . . . no

    We have: . (x^2 - 4x) - 4(y^2 + 2y) \:=\:16

    Complete the square: . (x^2 - 4x {\color{blue} \:+\: 4}) - 4(y^2 + 2y {\color{red}\:+\: 1}) \;=\;16 {\color{blue}\:+\: 4}  {\color{red}\:-\: 4}

    And we have: . (x-2)^2 - 4(y+1)^2 \:=\:16

    Divide by 16: . \boxed{\frac{(x-2)^2}{16} - \frac{(y+1)^2}{4} \;=\;1}



    To find the asymptotes, set the equation equal to 0.

    . . \frac{(x-2)^2}{16} - \frac{(y+1)^2}{4} \:=\:0 \quad\Rightarrow\quad \frac{(y+1)^2}{4} \;=\;\frac{(x-2)^2}{16} \quad\Rightarrow\quad (y\;+\;1)^2 \;=\;\frac{(x-2)^2}{4}

    Take square roots: . y + 1 \;=\;\pm\tfrac{1}{2}(x-2)


    Therefore: . y \:=\:\pm\tfrac{1}{2}(x-2) + 1\;\;\Rightarrow\;\;\begin{Bmatrix}y \:=\:\frac{1}{2}x - 2 \\ y \:=\:-\frac{1}{2}x \end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    what about my second question?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,520
    Thanks
    1404
    Quote Originally Posted by Paymemoney View Post
    Hi
    can someone show me how would you find the equation express in the form \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1 and asymptotes of the following equations:

    1) x^2-4y^2-4x-8y-16=0
    i got \frac{(x-2)^2}{4}-(y-1)^2=6


    2) 4x^2-8x-y^2+2y=0
    This is what i have done:
    4(x^2-2x)-(y-2y)=0

    (x-1)^2-\frac{(y-1)^2}{4}=0

    answer is \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1

    P.S
    2) 4x^2 - 8x - y^2 + 2y = 0

    4(x^2 - 2x) - (y^2 - 2y) = 0

    4(x^2 - 2x + 1) - (y^2 - 2y + 1) = 0 + 4 - 1

    4(x - 1)^2 - (y - 1)^2 = 3

    \frac{4(x - 1)^2}{3} - \frac{(y - 1)^2}{3} = 1.


    Now to find the asymptotes, set the equation equal to 0.

    So \frac{4(x - 1)^2}{3} - \frac{(y - 1)^2}{3} = 0

    \frac{(y - 1)^2}{3} = \frac{4(x - 1)^2}{3}

    (y - 1)^2 = 4(x - 1)^2

    y - 1 = \pm 2(x - 1)

    y = 1 \pm 2(x - 1).


    Therefore y = 2x - 1 or y = -2x + 3.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by Prove It View Post
    2)
    4(x^2 - 2x + 1) - (y^2 - 2y + 1) = 0 + 4 - 1
    can you explain to me how did you get 0 + 4 - 1??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,520
    Thanks
    1404
    Quote Originally Posted by Paymemoney View Post
    can you explain to me how did you get 0 + 4 - 1??
    To complete the square in x I had to add 4 to both sides and to complete the square in y I had to subtract 1 from both sides.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2008
    Posts
    509
    can someone tell me where i have gone wrong in finding the x-intercepts for question 1.

    \frac{4(x-1)^2}{3}-{(0-1)^2}{3}=1
    \frac{4(x-1)^2}{3}=1+\frac{1}{3}
    4(x-1)^2=\frac{4}{3}*3
    (x-1)^2=\frac{12}{3}*\frac{1}{4}
    x=\sqrt1+1
    so x=0 or x=2
    i looked in the book's answers and it says 1-\frac{\sqrt3}{2} and 1+\frac{\sqrt3}{2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,520
    Thanks
    1404
    Quote Originally Posted by Paymemoney View Post
    Hi
    can someone show me how would you find the equation express in the form \frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1 and asymptotes of the following equations:

    1) x^2-4y^2-4x-8y-16=0
    i got \frac{(x-2)^2}{4}-(y-1)^2=6


    2) 4x^2-8x-y^2+2y=0
    This is what i have done:
    4(x^2-2x)-(y-2y)=0

    (x-1)^2-\frac{(y-1)^2}{4}=0

    answer is \frac{4(x-1)^2}{3}-\frac{(y-1)^2}{3}=1

    P.S
    The x intercepts are where y = 0.

    So x^2-4y^2-4x-8y-16=0

    x^2 - 4x - 16 = 0

    x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-16)}}{2(1)}

     = \frac{4 \pm \sqrt{16 + 64}}{2}

     = \frac{4 \pm \sqrt{80}}{2}

     = \frac{4 \pm 4\sqrt{5}}{2}

     = 2 \pm 2\sqrt{5}.


    I do not agree with either answer...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Dec 2008
    Posts
    509
    oops its question 2 which i am having trouble finding the x-intercepts
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,520
    Thanks
    1404
    4x^2 - 8x - y^2 + 2y = 0


    Let y = 0, then

    4x^2 - 8x = 0

    4x(x - 2) = 0

    4x = 0 or x - 2 = 0

    x = 0 or x = 2.


    I agree with your answer...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok thanks that means the book's answer is incorrect.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rectanglar Hyperbola Proof
    Posted in the Geometry Forum
    Replies: 0
    Last Post: April 17th 2010, 06:02 AM
  2. Hyperbola question
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: May 26th 2009, 12:12 PM
  3. Hyperbola question and Log
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 9th 2009, 04:26 AM
  4. A small hyperbola question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 12th 2009, 12:01 AM
  5. Hyperbola Short Question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 21st 2007, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum