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Thread: Two parabolas intersecting to form rectangle

  1. #1
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    Two parabolas intersecting to form rectangle

    Suppose I have two parabolas defined by the following equations
    $\displaystyle f_1 = -x^2 + bx + c$ which has vertex $\displaystyle V_1 = (\frac{b}{2} , c + \frac{b^2}{4})$
    $\displaystyle f_2 = x^2$ which has vertex $\displaystyle V_2 = (0,0)$

    For some values $\displaystyle b, c$, these parabolas intersect as points $\displaystyle R, S$

    The first question asks
    (i) Determine all pairs $\displaystyle (b,c)$ for which the points $\displaystyle R, S$ exist and the points $\displaystyle V_1, V_2, R, S$ are distinct.
    Setting $\displaystyle f_1 = f_2$ results in $\displaystyle 2x^2- bx - c = 0$, and two distinct solutions x exist when the discriminant of the quadratic is greater than 0.

    Thus, it's all pairs $\displaystyle (b,c)$ such that $\displaystyle D = b^2 + 8c > 0$

    (ii) Determine all pairs $\displaystyle (b,c)$ for which the points R and S exist, the points $\displaystyle V_1,V_2,R,S$ are distinct, and quadrilateral $\displaystyle V_1RV_2S$ is a rectangle.

    This is where I'm a bit stumped. I think I would still have a range of solutions as (i) and that I would use some of these properties of rectangles.
    * Opposite sides have equal slope
    * Adjacent sides have negative reciprocal slope
    * The intersection of diagonals occurs at the midpoint.
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  2. #2
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    Re: Two parabolas intersecting to form rectangle

    I figured out a way to solve it. You can calculate R and S in general form using the quadratic formula, then squaring the x values to get the y values. You can then use the fact that rectangle diagonals are equal.
    Thus d(V_3, V_4) = d(R,S). Thus you'll have an equation that b and c must satisfy for the resulting shape $\displaystyle V_1RV_2S$ to be a rectangle.
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  3. #3
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    Re: Two parabolas intersecting to form rectangle

    (i) we also need $\displaystyle c\neq 0$ to make sure we have four distinct points

    (ii) interesting
    Thanks from MacstersUndead
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