Math

a parabola whose rootsare at A(-3,0) and B(6,0) has a y-intercept at point C(0,p),p<0, such that angle ABC is a right angle. if the equation of the parabola is y=ax^2+bx=c, compute the value of (abc)^2

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asked by Stan
  1. The roots of the parabola are known (-3 and 6), so the equation of the parabola is in the form:
    k(x+3)(x-6)=0 or
    k(x²-3x-18)=0 ...(1)
    where k is a constant to be determined.

    Since the y-intercept is negative, k>0.

    If ∠ABC is a right angle, then a circle with AB as diameter intersects the y-axis at intercepts (0,p) and (0,-p) where p<0.

    The centre of circle ABC is at ((6-3)/2,0), or (1.5,0), and the radius is (6-(-3))/2 = 4.5.

    The equation of circle ABC is therefore:
    C1 : (x-1.5)^2+y^2=4.5^2, or
    C1: (x-1.5)^2 + y^2 = 20.25

    To find intersection with the y-axis, substitute x=0 in C1 and solve for y to get
    y=±√18=±3√2

    So p=-3√2.

    The constant term of equation (1) should equal p (both are y-intercepts), which leads to:
    -3√2 = -18k

    Solve for k and substitute in (1) to get the equation of the parabola, and hence calculate (abc)^2 as required.

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