Results 1 to 4 of 4

Math Help - [SOLVED] Equation into standard form of a conic

  1. #1
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52

    [SOLVED] Equation into standard form of a conic

    I am trying to put the equations into standard form of a conic section...
    But cannot find its factors properly i have those three equations left
    Show all steps please...
    Thanks


    1. 3x^2-y^2+9x+8y-145/4=0

    2. 4x^2+9y^2+24x+18y+12=0

    3. y^2+10y=2x-1


    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,894
    Thanks
    326
    Awards
    1
    Quote Originally Posted by xterminal01 View Post
    I am trying to put the equations into standard form of a conic section...
    But cannot find its factors properly i have those three equations left
    Show all steps please...
    Thanks


    3x^2-y^2+9x+8y-145/4=0
    3x^2 - y^2 + 9x + 8y -\frac{145}{4} = 0

    Let's get rid of that nasty fraction:
    12x^2 - 4y^2 + 36x + 32y - 145 = 0

    Complete the square for x and y:
    (12x^2 + 36x) + (-4y^2 + 32y) = 145

    12(x^2 + 3x) - 4(y^2 - 8y) = 145

    12 \left ( x^2 + 3x + \frac{9}{4} \right ) - 4 (y^2 - 8y + 16) = 145 + 12 \cdot \frac{9}{4} - 4 \cdot 16

    12 \left ( x + \frac{3}{2} \right ) ^2 - 4(y - 4)^2 = 108

    \frac{ \left ( x + \frac{3}{2} \right ) ^2}{\frac{108}{12}} - \frac{(y - 4)^2}{\frac{108}{4}} = 1

    \frac{ \left ( x + \frac{3}{2} \right ) ^2}{3^2} - \frac{(y - 4)^2}{(\sqrt{27})^2} = 1
    which is the form for a hyperbola.

    The rest are done in a similar fashion.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52
    Thank you I appreciate it,
    I have tried to do the # 2 and 3 myself and this is what i came up with

    2. 4x^2+9y^2+24x+18y+12=0

    4x^2+9y^2+24x+18y=-12
    4(x^2+6x+9)+9(y^2+2y+1)=-12+36+9

    4(x+3)^2 + 9(y+1)^2 = 33

    (x+3)^2 + (y+1)^2
    33 33 = 1
    4 9


    and # 3 i guess is a parabola and this is what i got

    y^2+10y=2x-1
    (y^2+10y+25) = 2x-1+25
    (y+5)^2 = 2x + 24
    (y+5)^2 = 2(x+12)

    Please let me know if i did the rest correct
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by xterminal01 View Post
    Thank you I appreciate it,
    I have tried to do the # 2 and 3 myself and this is what i came up with

    2. 4x^2+9y^2+24x+18y+12=0

    4x^2+9y^2+24x+18y=-12
    4(x^2+6x+9)+9(y^2+2y+1)=-12+36+9

    4(x+3)^2 + 9(y+1)^2 = 33

    (x+3)^2 + (y+1)^2
    33 33 = 1
    4 9
    (how about some squares as denominator?)

    and # 3 i guess is a parabola and this is what i got

    y^2+10y=2x-1
    (y^2+10y+25) = 2x-1+25
    (y+5)^2 = 2x + 24
    (y+5)^2 = 2(x+12)

    Please let me know if i did the rest correct
    Hi,

    your calculations are OK

    Only with #2 I have added a (maybe necessary) remark

    (And now it's time to learn a little bit L_A T^E \chi)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Putting the equation of a conic into standard form
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 24th 2011, 08:28 PM
  2. Reduce the conic to standard form
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 7th 2010, 05:09 AM
  3. conic standard form/conversion
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 17th 2010, 01:33 PM
  4. Replies: 1
    Last Post: February 16th 2010, 07:21 AM
  5. math/check+ help! conic in standard form
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 5th 2006, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum