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Math Help - hyperbola and conic rotation

  1. #1
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    hyperbola and conic rotation

    Alright so im having trouble with 5 problems dealing with hyprebola, and conic rotation If someone can help me that would be great thanks

    Question 1
    find the canter, vertices, foci and the equations of the asymptotes of the hyperbla and sketch the graph. of x^2-9y^2+36y-72=0

    Question 2
    Find the standard form of the equation of the hyperbola
    Vertices(2,3),(2,-3); Foci (2,6)(2,-6)

    Question 3
    Find the standard form of the equation of the hyperbola
    Vertices(3,0)(3,4); Asymptotes: y=(2/3)x, y=4-(2/3)x

    Question 4 (this is dealing with conic rotation i dont know how to do this at all so if someone can explain it a bit that would be nice)
    use the discriminant to classify the graph. Then use the quadratic formula to solve for y and use a graphing utility to graph the equation
    x^2-8xy+y^2-10x+5y=0

    Question 5
    Rotate the axes to eliminate the xy-term. Sketch the graph of the resulting equation, showing both sets of axes.
    xy-2=0
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  2. #2
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    Hello, ladventchildp!

    I'll walk through the first one . . .


    1) Find the center, vertices, foci and the equations of the asymptotes
    and sketch the graph: .x - 9y + 36y - 72 .= .0

    Get the equation in standard form . . .

    We have: .x - 9(y - 4y . . .) .= .72

    Then: . . . x - 9(y - 4y + 4) .= .72 - 36

    . . . . . . . . . . . x - 9(y - 2) .= .36

    . . . . . . . . . . . x . (y-2)
    Divide by 36: . --- - ------- .= .1
    . . . . . . . . . . .36 . . .4


    The hyperbola has center C(0,2) . . . It opens left and right.
    . . We have: .a = 6, b = 2

    The vertices are: .(6, 2)

    The "focus equation" is: .c .= .a + b
    . . So we have: .c .= .36 + 4 .= .40 . . . . c .= .2√10
    The foci are: .(2√10, 2)

    The asymptotes are: .y .= .(1/3}x

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  3. #3
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    Hello, ladventchildp!

    If you are expected to do Rotation problems,
    . . you should have been given a lot more information and several formulas.


    4) Use the discriminant to classify the graph. .Then use the Quadratic Rormula
    to solve for y and use a graphing utility to graph the equation.
    . . x - 8xy + y - 10x + 5y .= .0
    The general quadratic equation is: .Ax + Bxy + Cy + Dx + Ey + F .= .0

    The discriminant is: .d .= .B - 4AC

    . . If d = 0: parabola
    . . . .d > 0: hyperbola
    . . . .d < 0: ellipse


    We have: .A = 1, B = -8, C = 1, D = -10, E = 5, F = 0

    The discriminant is: .d .= .(-8) - 4(1)(1) .= .+60 . . . positive: hyperbola


    Write the equation in terms of y:

    . . y - (8x - 5)y + (x - 10x) .= .0
    . . . - . . . . - . . . . . . . . . . . . . . . . . . _____________________
    . . . . . . . . . . . . . . . . . . . (8x - 5) √(8x - 5) - 4(1)(x - 10x)
    Quadratic Formula: . y .= . --------------------------------------------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2(1)

    . . . . . . . . . . . . . . . . . . . . . . . . ______________
    . . . . . . . . . . . . . . . (8x - 5) √60x - 40x + 25
    And we have: . y .= .--------------------------------
    . . . . . . . . . . . . . . . . . . . . . . . .2



    5) Rotate the axes to eliminate the xy-term.
    Sketch the graph of the resulting equation,
    showing both sets of axes: .xy - 2 .= .0
    We have: .A = 0, B = 1, C = 0, D = 0, E = 0, F = -2

    The discriminant is: .d .= .1 - 4(0)(0) .= .+1 . . . positive: hyperbola

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B
    The angle of rotation (θ) is given by: . tan(2θ) .= .-------
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C
    . . . . . . . . . . . . . . . . . . . . 1
    . . We have: . tan(2θ) .= . ------- .= .
    . . . . . . . . . . . . . . . . . . . 0 - 0

    Hence: . .= .π/2 . . . . θ .= .π/4

    . . The hyperbola has been rotated 45 CCW.


    How do we "un-rotate" it?

    There is another set of formulas:
    . . x' .= .xcosθ - ysinθ
    . . y' .= .xsinθ + ycosθ

    . . . . . . . . . . . . . . . . . . . . . . . . ._ . - . _ . . - . - . . . _ . . . . _
    Since θ = 45, we have: .x' .= .x/√2 - y/√2, . y' .= .x/√2 + y/√2


    Substitute into the original equation: .xy .= .2
    . . . . . . . . . . . _ . . . ._ . . . _ . . . . _
    We have: . (x/√2 - y/√2)(x/√2 + y/√2) .= .2

    . . . . . . . . . . . . . . . . . . . .x - y .= .2

    Therefore, we have: . . . . . . . . x - y .= .4

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