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Rotation of Axes showing parabola

  1. May 28, 2012 #1
    I was given the equation: 2sqrt(2)(x+y)^2=7x+9y

    I need to then:
    a)Use rotation of axes to show that the following equation represents a parabola
    b) Find the XY- and xy- coordinates of the vertex and focus
    c) Find the equation of the directrix in XY- and xy-coordinates


    Formulas provided:

    General Equation of a Conic
    Ax^2+Bxy+Cy^2+Dx+Ey+F=0

    cot2(fi) =A-C/B

    x=Xcos(fi)-Ysin(fi)
    y=Xsin(fi)+Ycos(fi)


    So, I got up to a certain point but now I am unsure as to what I must do.
    I solved for fi, which equals 45 degrees

    I solved for x and y which is

    x= X/sqrt(2) -Y/sqrt(2)
    y= X/sqrt(2)+Y/sqrt(2)


    I think I should plug it into the original equation to get x^2=4py or y^2=4px?
    Then I am just oblivious as to how I should solve for directrix and vertex,focus for XY, and xy

    Any help would be appreciated
    This section is new to me so I am working on learning the fundamentals of this section and this touches upon all aspects of the section.
     
  2. jcsd
  3. May 28, 2012 #2

    tiny-tim

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    welcome to pf!

    hi indio1! welcome to pf! :smile:

    (have a square-root: √ and a phi: φ and try using the X2 button just above the Reply box :wink:)
    that's right! :smile:

    you should get something like X2 + Ax + By = 0

     
  4. May 28, 2012 #3
    So, i then get this

    4x'2 - 8x' - y' = 0.

    Put in other form:

    y' = 4x'2 - 8x'

    y' + 4 = 4(x'2 - 2x' + 1)

    y' = 4(x' - 1)2 - 4

    vertex is at (1,-4) in x'y'-coordinates. use the formulae for expressing x,y in terms of x',y'.

    the x-coordinate of the vertex is (1)(√2/2) - (-4)(√2/2) = (5√2)/2
    the y-coordinate of the vertex is (1)(√2/2) + (-4)(√2/2) = -(3√2)/2

    standard form:

    (x' - 1)2 = 4(1/16)(y' + 4)

    focus of the parabola in x'y'-coordinates is at (1,-63/16) and the directrix is the line y' = -65/16.



    (1)(√2/2) - (-63/16)(√2/2) = (79√2)/32
    (1)(√2/2) + (-63/16)(√2/2) = -(47√2)/32



    line is of the form: y - y0 = x - x0.


    (0)(√2/2) - (-65/16)(√2/2) = (65√2)/32 <---x0
    (0)(√2/2) + (-65/16)(√2/2) = -(65√2)/32 <---y0.

    this tells us the equation of the directrix in xy-coordinates is: y = x - (65√2)/16

    According to wolframalpha, something is wrong here.
    the focus is at the point 79/16sqrt(2) and -47/16sqrt(2) and the directrix is at the point x -65/8sqrt(2)


    What is wrong here?
     
  5. May 28, 2012 #4

    tiny-tim

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    looks fine down to here :smile:

    (i can't remember anything about focus and directrix, so i can't comment on the rest :redface:)
     
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