Results 1 to 4 of 4

Math Help - I Need Help With Ellipses!!! And Conics!!!

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    2

    I Need Help With Ellipses!!! And Conics!!!

    Hi. I'm in a panic, because i have this take home quiz i need to do for class, and I'm pretty lost when it comes to algebra II.

    The homework says that I need to scetch the ellipse whose foci are (1,4) and (1,2)
    and whose minor axis is 8 units, then i have to write the equation of that ellipse.

    also i need to, from standard form, classify each conic, the identify its center:

    1. (x+6)^2 ---(y-5)^2
    ________ + _________= 1
    ------- 81 - - - - - 64

    2. x^2+8y^2-4x-16y-4=0

    3. x^2+y^2+6x+6y=0

    And of course, I forgot my textbook in my locker, and I've searched all over the web, but am still having trouble understanding this stuff. Someone please help me.


    -Tessa
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by blueskiiiedchica View Post
    [snip]
    i need to, from standard form, classify each conic, the identify its center:

    1. (x+6)^2 ---(y-5)^2
    ________ + _________= 1
    ------- 81 - - - - - 64

    2. x^2+8y^2-4x-16y-4=0

    3. x^2+y^2+6x+6y=0

    And of course, I forgot my textbook in my locker, and I've searched all over the web, but am still having trouble understanding this stuff. Someone please help me.


    -Tessa
    How long have you had to prepare for this take-home quiz. You need to do a bit more work and get better organised.

    Standard form of ellipse with centre at (h, k): \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.

    1. See above.

    2 and 3. Complete the square in x and y and re-arrange into above standard form.

    Eg. x^2 + 6y^2 - 4x + 12y - 15 = 0 \Rightarrow (x^2 - 4x) + (6y^2 + 12 y) - 15 = 0

     \Rightarrow (x - 2)^2 - 4 + 6 (y^2 + 2y) - 15 = 0 \Rightarrow (x - 2)^2 - 4 + 6[(y + 1)^2 - 1] - 15 = 0

    \Rightarrow (x - 2)^2 - 4 + 6(y + 1)^2 - 6 - 15 = 0 \Rightarrow (x - 2)^2 - 4 + 6[(y + 1)^2 - 1] - 15 = 0

    \Rightarrow (x - 2)^2 - 4 + 6(y + 1)^2 - 6 - 15 = 0 \Rightarrow (x - 2)^2 + 6(y + 1)^2 = 25

     \Rightarrow \frac{(x - 2)^2}{25} + \frac{6(y + 1)^2}{25} = 1 etc.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    2
    Oh, thank you so much! I've been scratching my head over this for hours. Yes, usually I would prepare better, but I've had bronchitis and have been a mess =[
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by blueskiiiedchica View Post
    Hi. I'm in a panic, because i have this take home quiz i need to do for class, and I'm pretty lost when it comes to algebra II.

    The homework says that I need to scetch the ellipse whose foci are (1,4) and (1,2)
    and whose minor axis is 8 units, then i have to write the equation of that ellipse.

    [snip]
    Label the focii on a set of axes. You should be able to draw a rough ellipse around them.

    You know the centre is half-way between the focii, right? So you can get h and k.

    And you know that the focii lie on the major axis, which in the case is parallel to the y-axis. So the minor axis is parallel to the x-axis. Therefore length of minor axis is 2a = 8 => a = 4.

    Use your diagram and Pythagoras' Theorem to see why b = \sqrt{17}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Ellipses
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 7th 2010, 05:15 PM
  2. Ellipses
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 30th 2009, 07:48 AM
  3. Conics - Ellipses + tangents
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 5th 2009, 05:15 AM
  4. Ellipses
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: November 13th 2007, 06:41 PM
  5. Replies: 2
    Last Post: August 28th 2007, 01:55 PM

Search Tags


/mathhelpforum @mathhelpforum