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Math Help - alg 2 conics

  1. #1
    Junior Member mattballer082's Avatar
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    alg 2 conics

    if possible help me step by step. i need to understand this.

    Find the equation of the hyperbola described:

    (a) foci (0, +-3) asymptotes y = +- 3/4

    (b) center (-1, 3) vertices (-4,3) and (2,3) foci (-6,3) and (4,3)..

    very confused. anyone?

    also how is this problem done with a circle and a parabola?

    the circle X (squared) + Y (squared) = 4 and parabola 5y = -3x (squared). intersect at the points:????
    how do i find the points?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mattballer082 View Post
    Find the equation of the hyperbola described:

    (a) foci (0, +-3) asymptotes y = +- 3/4
    The asymptotes indicate that this is not a hyperbola. Is there a typo here?

    Quote Originally Posted by mattballer082 View Post
    also how is this problem done with a circle and a parabola?

    the circle X (squared) + Y (squared) = 4 and parabola 5y = -3x (squared). intersect at the points:????
    how do i find the points?
    x^2 + y^2 = 4
    5y = -3x^2

    Typically we would solve the bottom equation for x or y and plug it into the top equation. But the bottom equation is practically solved anyway, so no trouble there:
    5y = -3x^2 --> y = -(3/5)x^2

    Insert this into the circle equation:
    x^2 + y^2 = 4

    x^2 + [-(3/5)x^2]^2 = 4

    Now solve for x:
    x^2 + 9/25 * x^4 = 4

    9/25 * x^4 + x^2 - 4 = 0 <-- Multiply both sides by 25

    9x^4 + 25x^2 - 100 = 0

    This is what is called a "biquadratic equation." Note that only even powers of x show up. I'm going to define
    z = x^2

    Then
    9z^2 + 25z - 100 = 0

    z = [-25 (+/-) sqrt{25^2 - 4*9*(-100)}]/(2*9)

    z = [-25 (+/-) sqrt{625 + 3600}]/18

    z = [-25 (+/-) sqrt{4225}]/18

    z = [-25 (+/-) sqrt{4225}]/18

    z = [-25 (+/-) 65]/18

    So z = 40/18 = 20/9 or z = -90/18 = -5.

    Thus
    z = x^2 implies that:
    x = (+/-)sqrt{20/9} or x = (+/-)sqrt{-5}

    So
    x = (+/-)(2/3)*sqrt{5}
    are the only two x values. (I'm ignoring the z = -5 solution because we can't take the square root of a negative number.)


    So what are the corresponding y values? Well:
    y = -(3/5)x^2

    y = -(3/5)*[(+/-)(2/3)*sqrt{5}] = -(3/5)(20/9) = -4/3 (for both the + and - solutions, of course.)

    So there are two intersection points:
    (x, y) = (2/3*sqrt{5}, -4/3)
    and
    (x, y) = (-2/3*sqrt{5}, -4/3)

    I have attached a graph of both curves below.

    -Dan
    Attached Thumbnails Attached Thumbnails alg 2 conics-intersection.jpg  
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