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Math Help - Parabola Tangent

  1. #1
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    Parabola Tangent

    You are given a parabola (1 - x^2). You have to find points (P and Q( on the parabola so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equliateral triangle.
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  2. #2
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    Hello, BarlowBarlow1!

    You are given a parabola: .y .= .1 - x
    You have to find points P and Q on the parabola so that the triangle ABC
    formed by the x-axis and the tangent lines at P and Q is an equliateral triangle.
    Code:
                          |
                         A*
                         /|\
                        / | \
                       /  |  \
                      /   *   \
                     / *  |  * \
                    /*    |    *\
                  P*      |      *Q
                  /       |       \
            -----/*-------+-------*\----
                B         |         C

    Let point P be (p, 1-p) .and Q be (q, 1-q)

    We know that the angles of an equilateral triangle are all 60.
    . . . - . . . - . . . - . . . . . . . . . . . . . . . . ._
    Hence, the slope of BA is: .tan(60) .= .√3


    We also know that the slope of a tangent is: .y' .= .-2x
    . . Hence, at point P(p, 1-p), the slope is: .-2p

    . - . . - . . - . . . . . . . . _ . - . . . - . . . . . . ._
    So we have: . -2p .= .√3 . . . . p .= .-√3
    . - . . - . . . . . . . . . . . ._
    By symmetry: .q .= .√3

    . - . . - . . - . . - . . . . ._ . . . . . . . . . . . _
    The points are: .P(-√3, ) .and .Q(√3, )

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