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Help with rotation of axes?

~Rewrite the equation in a rotated x'y'-system without an x'y' term. Express the equation involving x' and y' in the standard form of a conic section.

24xy - 7y^2 + 36 = 0

My teacher did a poor job of explaining this section and I have no clue what to do! Any help will be greatly appreciated! :)

1 Answer

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  • 7 years ago
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    If A is the rotation angle of the new co-ordinate system, we can express the current x, y co-ordinates in terms of the new co-ordinate system as:

    x = x' cos A - y' sin A

    y = x' sin A + y' cos A

    Substituting these values into 24xy - 7y^2 + 36 = 0, we get:

    24 (x' cos A - y' sin A)(x' sin A + y' cos A)

    - 7 (x' sin A + y' cos A)^2 + 36

    = 24 [x'^2 (sin A) (cos A) + (x'y')(cos^2 A - sin^2 A) - (y'^2) (sin A) (cos A)]

    - 7 [x'^2 sin^2 A + (2x'y')(sin A) (cos A) + (y'^2 cos^2 A)

    + 36

    Collecting x'^2, x'y' and y'^2 terms:

    = x'^2 [24 (sin A) (cos A) - 7 sin^2 A]

    + x'y'[(cos^2 A - sin^2 A) - 14(sin A) (cos A)]

    + y'^2 [-24 (sin A) (cos A) - 7 cos^2 A]

    + 36

    But 2 (sin A) (cos A) = sin 2A,

    so 24 (sin A) (cos A) = 12 sin 2A

    Also cos 2A

    = cos^2 A - sin^2 A

    = 1 - sin^2 A - sin^2 A = 1 - 2 sin^2 A

    so 1 - cos 2A = 2 sin^2 A

    So - 7 sin^2 A = (-7/2)(1 - cos 2A)

    so the coefficient of x'^2 is 12 sin 2A -(7/2)(1 - cos 2A)

    Since cos 2A = cos^2 A - sin^2 A

    = cos^2 A - (1 - cos^2 A)

    = 2 cos^2 A - 1,

    then 1 + cos 2A = 2 cos^2 A,

    so -(7/2)(1 + cos 2A) = - 7 cos^2 A,

    Then coefficient of y'^2 is:

    -24 (sin A) (cos A) - 7 cos^2 A

    = -12 sin 2A - (7/2)(1 + cos 2A)

    The coefficient of x'y' is therefore

    cos 2A - 7 sin 2A.

    To easily identify the conic section, we want this coefficient to vanish, so we choose A accordingly. If we then divide by cos 2A, we get:

    1 - 7 tan 2A = 0, so tan 2A = 1,

    i.e. tan 2A = 1/7.

    The equation in the new co-ordinate system becomes:

    [12 sin 2A -(7/2)(1 - cos 2A)] x'^2

    + [0] x'y'

    + [-12 sin 2A - (7/2)(1 + cos 2A)] y'^2

    + 36 = 0

    Dividing through by cos 2A:

    [12 tan 2A -(7/2)(sec 2A - 1)]x'^2

    + [-12 tan 2A - (7/2)(sec 2A + 1)] y'^2

    + 36 sec 2A = 0

    But tan 2A = 1/7,

    and sec 2A = sqrt (sin^2 2A + cos^ 2A) / (cos 2A)

    = sqrt (tan^2 2A + 1)

    = sqrt (1/49 + 1) = sqrt (50/49) = sqrt(50)/7

    So:

    [(12/7) +(7/2)(1 - sqrt(50)/7)]x'^2

    - [(12/7) + (7/2)(1 + sqrt(50)/7] y'^2

    + 36 sqrt(50)/7 = 0

    Multiplying by 14:

    [24 +7(7 - sqrt(50))]x'^2

    - [24 + 7(7 + sqrt(50)] y'^2

    + 72 sqrt(50) = 0

    =[24+49-7 sqrt(50)]x'^2

    - [24 + 49+7 sqrt(50)] y'^2

    + 72 sqrt(50)

    =[73-7 sqrt(50)]x'^2

    - [73+7 sqrt(50)] y'^2

    + 72 sqrt(50)

    Define u=73-7 sqrt(50), v=73+7 sqrt(50), w=72 sqrt(50).

    Then this can be expressed as:

    u x'^2 - v y'^2 + w = 0.

    Dividing through by w and rearranging:

    1 = (v/w) y'^2 - (u/w) x'^2.

    This is consistent with the form:

    y'^2/a^2 - x'^2/b^2 = 1

    which is a hyperbola.

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