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Conic section equation

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    I have : (x)^2+(2xy)+y^2+2(2)^(1/2)x−2(2)^(1/2)y+4=0

    Now when I use the rotation formulas I get : y'=(x'^2)/2 + 1

    But I don't know if this makes sense....

    2. Relevant equations


    tan2β=B/(A−C)

    x=x'cos β - y'sin β

    y=x'cos β + y'sin β
    3. The attempt at a solution

    We know that β= pi/4 because B/(A−C) tends towards infinity.

    So we replace pi/4 in
    x=x'cos β - y'sin β

    y=x'cos β + y'sin β

    and put those in the long equation and simplify. Which gives us y'=((x'2)/2)+1

    Does this make sense my answer
     
    Last edited: Nov 9, 2015
  2. jcsd
  3. Nov 9, 2015 #2

    Mark44

    Staff: Mentor

    Right. What is x'2 supposed to mean?
     
  4. Nov 9, 2015 #3
    Oh sorry, what I meant is (x'^2)/2 + 1
     
  5. Nov 9, 2015 #4

    Mark44

    Staff: Mentor

    This is extremely hard to read. It would be much better to write it using LaTeX, which this forum supports.

    ##x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0##

    What I wrote was ##x^2 + 2xy + y^2 + 2\sqrt{2}x - 2\sqrt{2}y + 4 = 0##
    We have a primer on how to use LaTeX (under INFO on the menu bar) -- https://www.physicsforums.com/help/latexhelp/
    Note: Corrected by OP in a later post to ##y' = (1/2)(x')^2 + 1##
    I don't think this is right. Your conic section is a circle, not a parabola.
     
    Last edited: Nov 9, 2015
  6. Nov 9, 2015 #5
    According to wolfram Im supposed to have a parabola... I dont know what went wrong tbh
     
  7. Nov 9, 2015 #6

    Mark44

    Staff: Mentor

    I don't know what you entered into WA, so I can't say why WA is telling you that your equation represents a parabola. The presence of the ##x^2## and ##y^2## terms, each with a coefficient of 1, says that this is a circle. The xy term indicates that there is a rotation. You can't rotate a circle to become a parabola.

    Your work on finding the angle of rotation looks OK to me. Show what you did for this part:
     
  8. Nov 9, 2015 #7
    Sorry for my late reply : http://www.wolframalpha.com/input/?i=x^2+2xy+y^2+2(2)^(1/2)x-2(2)^(1/2)y+4=0

    Also there is now way this can be a circle because my exercice makes it clear that its an ellipse, hyperbola or parabola
     
  9. Nov 9, 2015 #8

    Mark44

    Staff: Mentor

  10. Nov 9, 2015 #9
    MARK this problem is going to make me insane. I'm pretty sure I applied everything correctly.
     
  11. Nov 9, 2015 #10

    Mark44

    Staff: Mentor

    The equation you wrote, in the transformed coordinates, is a parabola, which agrees with what WA is showing. Is there still something you don't understand?
     
  12. Nov 9, 2015 #11
    Ok, if what I have is correct, why does wolfram show it differently in this case : http://www.wolframalpha.com/input/?i=+y=((x^2)/2)+1

    This looks different from the other parabola..

    Edit: Oh right I rotate it by pi/4, so I guess that I must do this myself right ? I just have no IDEA on how to rotate by graph.
     
  13. Nov 9, 2015 #12

    Mark44

    Staff: Mentor

    Think about it this way: instead of rotating the graph by ##\pi/4## counterclockwise, rotate the axes ##\pi/4## clockwise.
     
  14. Nov 9, 2015 #13
    The problem isn't the rotating, the problem is how do I graph a rotated parabola ? I have no idea on how you do this becaus Ive never done it. How do I even begin ? And btw I Must rotate it counter clock wise because my exercice offers me only the second quadrant. How do I find the vertex and the focus?
     
  15. Nov 9, 2015 #14

    Mark44

    Staff: Mentor

    The centerline of the parabola lies along the line y = -x. The vertex of the parabola in (x', y') coordinates is at (0, 1). If this point is rotated by ##\pi/4## where does it go? Two other points in the (x', y') coordinates are at (1, 3/2) and (-1, 3/2). Figure out where these points rotate to. Then you have three points and the centerline in (x, y) coordinates, so you should be able to get a rough graph of the parabola.
     
  16. Nov 9, 2015 #15
    These two other point you took them randomly ? Also, I need to determine what is the focus in this problem. How do u do that
     
  17. Nov 9, 2015 #16

    Mark44

    Staff: Mentor

    Not randomly. I used x' values of -1, 0, and 1, and calculated the y' values.

    To find the focus, see https://en.wikipedia.org/wiki/Parabola#Focal_length. You'll need to figure out where the focus is for y' = (1/2)(x')2, and then translate it up by 1 unit. Finally, rotate this point by ##\pi/4## to get the (x, y) coordinates.
     
  18. Nov 9, 2015 #17
    Ok, but one thing which I don't understand. When I find my normal point of x',y' and when I rotate these points, is there some analytical method which gives me the rotation of these points? Or I do all of this by drawing ?
     
  19. Nov 9, 2015 #18
    Ok I was able to find (0,3/2) for my focus and I was able to draw my parabola. Now it's the rotation which I need to figure out how to do.
     
  20. Nov 9, 2015 #19

    Mark44

    Staff: Mentor

    You can find them by drawing them, or you can identify the points using simple trig. For example, the vertex in (x', y') coordinates is at (0, 1). If you rotate this point CCW, you get ##(-\sqrt{2}/2, \sqrt{2}/2)##, or about (-.71, .71). The other points will be a little bit harder, but keep in mind that the radial distance (distance between the point and the origin) doesn't change.
     
  21. Nov 9, 2015 #20
    I hope that I don't bother you with my one thousand questions. First, can you tell me the analytic method ? Like some wiki page. Also, the radial distance is the distance between any point of the parabola and the origin right ?

    I need to show how I obtained my data so I need the analytical method.
     
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