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Identifying equations as parabola, hyperbolas, ellipses, circles, straight lines?

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data

    when asked to identify an equation as a parabola, hyperbolas, ellipses, circles, straight lines, or none of the above, how can i deduce which is which?

    the problems given are:

    2x^2+2y^2=9 (im pretty sure this one's a circle, just by graphing it, but id like confirmation)
    x^2/16 + y/25 =1
    3x^2 =7 + 3y^2
    x/16 + y/25 = 1
    x^2 = 16 - (y-3)^2
  2. jcsd
  3. May 21, 2008 #2


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    Homework Helper

    That can be difficult, especially if the graph is plotted on axes which have been translated, rotated or have undergone some linear transformation. You have to express it in the form of a matrix, find some eigenvalues and eigenvectors and then re-express the entire matrix equation in terms of the new coordinate axes. It's difficult to explain here. What do your notes say?

    On the other hand, if they are plotted on the standard coordinate axes, then use this as a rough guide:

    Ellipse (a circle is a special type of ellipse):

    [tex]\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1[/tex]


    [tex]\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1 \ \mbox{or} \ -\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1 [/tex]


    [tex]x^2 = \alpha y \ \mbox{or} \ y^2 = \alpha x[/tex]

    [tex] \alpha[/tex] and [tex]\beta[/tex] are arbitrary non-zero real numbers.
  4. May 21, 2008 #3
    Lines can have the form

    y=mx+b or Ax+By+C=0

    For circles: coefficients are 1

    Best thing is to go thru ur book: Write down all the general forms you can find, and analyze the differences in each one. If you have all of them in front of you, it will be easier to find a pattern.
  5. May 21, 2008 #4
    ok heres what i came up with...
    2x^2+2y^2=9 CIRCLE
    x^2=16-4y^2 CIRLCE
    x^2/16 + y/25 =1 ELLIPSE
    3x^2 =7 + 3y^2 ELLIPSE
    x/16 + y/25 = 1 PARABOLA.

    did i get them right? i really need confirmation asap thanks for all the help guys!
  6. May 21, 2008 #5
    ALL WRONG but 1.

    1st one is correct, rest is wrong.

    Pay attention to whether your variable is LINEAR or QUADRATIC!!!
  7. May 21, 2008 #6
    im confused. can someone walk me through this step by step?

    i can keep guessing but i dont have much time and i need to figure this out. so can someone offer some guidance?

    thanks :)

    EDIT after looking through my textbook ive changed my answers

    2x^2+2y^2=9 CIRCLE
    x^2=16-4y^2 PARABOLA
    x^2/16 + y/25 =1 PARABOLA
    3x^2 =7 + 3y^2 HYPERBOLA
    x/16 + y/25 = 1 LINE
    x^2 = 16 - (y-3)^2 CIRCLE
    Last edited: May 21, 2008
  8. May 21, 2008 #7
    Did you type them in correctly? If so, then you only got the 1st one correct.

    Go back and read Defender's post!!! Compare your answers to the general forms he gave you.
  9. May 21, 2008 #8
    i did, and edited my answers. one thing i wasnt sure about was something like this:

    3x^2 =7 + 3y^2

    are we assuming u subtract 3y^2 from both sides so u get:

    3x^2 - 3y^2 =7

    which looks like the formula for a hyperbola? anyway, someone please check my answers. thanks!
    Last edited: May 21, 2008
  10. May 21, 2008 #9
    ONCE AGAIN, read Defender's post and compare it to yours.

    Before I sleep, "pay close attention" to the powers of your variables, this APPLIES to both x and y.

    If x and y are both linear, then it's a line. If x and y are both quadratic, then it's either a circle, ellipse, or hyperbola.

    If x or y is either linear or quadratic, vice versa, then it's a parabola.

    ... read your book and quit using your lame calculator, gnite and goodluck
    Last edited: May 21, 2008
  11. May 21, 2008 #10
    AH got it.

    x^2/16 + y/25 =1
    3x^2 =7 + 3y^2
    x/16 + y/25 = 1
    x^2 = 16 - (y-3)^2
  12. May 21, 2008 #11
    2nd one still wrong, rest good.
  13. May 21, 2008 #12
    how? it has 2 exponents meaning its a hyperbola, circle, or ellipse, and the denominators are the same making it a circle? right?
  14. May 21, 2008 #13


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    [tex]x^2 = 16 - 4y^2[/tex]

    The denominators are not the same. Yes, it has two square terms.

    Move all the terms to the LHS, and try to make the RHS 1. Now look at the possible conic section equations given above. Which type is your equation?
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