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Homework Help: Sketch polar curve (conic i believe)

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data


    Let C be the curve whose equation is given by: 4x^2+18y-20=9y^2+16x+9

    Identify and sketch the curve C and if possible give its center, directrix, foci, vertices, and endpoints on the minor axis.

    2. Relevant equations
    I have learned parabola, hyperbola and elispes (these are conic sections)



    3. The attempt at a solution

    To start with, I really can't. I tried to rearrange the equations.
    4x^2 - 9y^2+18y-16x=29
    but i don't see how i can graph this... please guide me through this, thank you.

    edited
    i just asked wolfram to do graph 4x^2 - 9y^2+18y-16x=29 and i got two hyperbolas (they are not symmetrics)... but how come? i don't see the 1 in any side
     
    Last edited: May 24, 2010
  2. jcsd
  3. May 24, 2010 #2

    Cyosis

    User Avatar
    Homework Helper

    Try to get rid of the first order x and y terms by completing the square. This way you can cast the equation into the form [itex]a(x-b)^2-c(y-d)^2=e[/itex], which should be familiar.
     
  4. May 24, 2010 #3
    so 4x^2+18y-20=9y^2+16x+9
    i get 4(x^2 - 4x + 4) - 9(x^2 + 2y +1) = 29
    so
    4(x-2)^2 - 9(x+1)^2 = 29

    is that correct?
    this is a parabola.

    but this 4(x-2)^2 - 9(x+1)^2 = 29 produces a different graph than the original form in wolfram.
    i did get a parabola, but using the original form i got something looked like hyperbola

    i got really confused because how can i tell what the equation supposes to be?
    like in the case 9x^2+y^2=35, the ans key said it was an ellipse....
     
    Last edited: May 24, 2010
  5. May 24, 2010 #4

    Mark44

    Staff: Mentor

    No, for several reasons.
    Your original equation involved both x and y. Your new equation does not. Also, when you completed the square, it looks like you forgot to add the same quantities to the right side. For example, when you added 4 inside the parentheses to complete the square in x^2 -4x, you are really adding 16. You need to add 16 to the right side also.
    That's because it is a hyperbola.
     
  6. May 24, 2010 #5

    Cyosis

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    Homework Helper

    You're interchanging x-es and y-s at random. That is why you get a parabola in your final expression. If you do it correctly you will get a hyperbola. You need to arrive at an equation of the form in post #2 which includes y.

    You should learn the 'definitions' of a hyperbola and an ellipse. I am pretty sure you will find them in your text book.

    Ellipse
    [tex]
    \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1
    [/tex]

    Hyperbola
    [tex]
    \frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}=1
    [/tex]

    Whenever you have a problem like this you check if you can cast your equation into one of the two forms above.
     
    Last edited: May 24, 2010
  7. May 24, 2010 #6
    oh my bad. it was a typo, no wonder why
    4(x-2)^2 - 9(y+1)^2 = 29
    now i have to get it back to the form
    with a^2 and b^2 as fraction and 1 as the difference
    how can i do that?
    our textbook only have examples with the regular form -_-
    do i just divide everything by 29? and so a = sqrt(29)/2 and b = sqrt(29)/3

    =_=
    (((x-2)^2)/(sqrt(29)/2)^2) - (((y+1)^2)/(sqrt(29)/3)^2) = 1

    it seems right to me in wolfram.... what a complicated form...
     
  8. May 24, 2010 #7

    Cyosis

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    Homework Helper

    Yes that is correct.
     
  9. May 24, 2010 #8

    Mark44

    Staff: Mentor

    I beg to differ. It is not correct.

    4x2 + 18y - 20 = 9y2 + 16x + 9
    <==> 4x2 - 16x - 9y2 + 18y = 29
    <==> 4(x2 - 4x + ?) - 9(y2 - 2y + ?) = 29

    Can you continue from here? There are at least two errors in your work - forgetting to change the sign of 18y when you factor out -9, and forgetting to add the right amounts to the right side after completing the square.

    When you're done, the number on the right side should NOT be 29.
     
  10. May 24, 2010 #9
    omg. i am so dead right now. i kept putting the wrong number. on the draft paper i did get -1 for the y

    oh i missed your second comment
     
  11. May 24, 2010 #10

    Mark44

    Staff: Mentor

    No, the 29 is wrong. There is no 29 in the final equation. In the last line of my previous post I had
    <==> 4(x2 - 4x + ?) - 9(y2 - 2y + ?) = 29

    To complete the square in the x terms, add 4. This means I am really adding 16 to the left side because of the multiplier 4 in front of the group, so I have to add 16 to the right side of the equation. When you complete the square in the y terms, you have to do something similar.

    If I were you I would go back and review completing the square.
     
  12. May 24, 2010 #11
    4(x^2 - 4x + 4) - 16 - 9(y2 - 2y + 1) - 9= 29 + 16 - 9
    am i getting 36?
     
  13. May 24, 2010 #12

    Mark44

    Staff: Mentor

    This is still wrong. THINK ABOUT WHAT YOU ARE DOING!

    On the left side you are adding 16 and then subtracting 16. You are also subtracting 9 and then subtracting 9 again.
     
  14. May 24, 2010 #13
    i thought the - carries to all the 9's there

    4(x^2 - 4x + 4) 16 - 9(y2 - 2y + 1) - 9= 29 + 16 + 9
    i really got confused here.

    when i did it in wolfram, complete square (the original form)
    i got this
    4 (x-2)^2-9 (y-1)^2-36 = 0
     
  15. May 24, 2010 #14

    Char. Limit

    User Avatar
    Gold Member

    Why would you add nine to the right side? The 9 on the left side is subtracted, not added...
     
  16. May 24, 2010 #15

    Mark44

    Staff: Mentor

    Well, yes, but why are you confused? This isn't complicated.

    4(x2 - 4x + ?) - 9(y2 - 2y + ?) = 29
    <==> 4(x2 - 4x + 4) - 9(y2 - 2y + ?) = 29 + 16
    I added 16 (= 4 * 4) to the left side, so to keep the equation equivalent to the one before it I added 16 to the right side. Can you work the rest of it by yourself (without Wolfram Alpha)?
    Until you understand completing the square, I would advise you not to use WA for this purpose. Besides, you are not likely to be able to use it for a test.
     
  17. May 24, 2010 #16
    Hi Mark
    I believe what you are saying is that whatever i added, i do the same on the other side
    whatever i substracted, i do the same on the other side
    since i added 16, i added 16 to 29
    i minused 9 so i -9

    am i correct there?
     
  18. May 24, 2010 #17

    Mark44

    Staff: Mentor

    Yes, that's exactly what I'm saying. Continuing the process from here -
    4(x2 - 4x + 4) - 9(y2 - 2y + ?) = 29 + 16

    what do you get?
     
  19. May 24, 2010 #18
    4(x2 - 4x + 4) - 9(y2 - 2y + 1) (-9) = 29 + 16 -9
    36?
     
  20. May 24, 2010 #19

    Mark44

    Staff: Mentor

    Why is this......................................^... here? I.e., (-9).
     
  21. May 24, 2010 #20
    i thought you agreed that
    4(x2 - 4x + 4) - 9(y2 - 2y + 1) (+1 * -9) = ....
    I am sorry if I still not able to answer the question .correctly ...
     
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