Math
posted by Stan on .
a parabola whose rootsare at A(3,0) and B(6,0) has a yintercept 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

The roots of the parabola are known (3 and 6), so the equation of the parabola is in the form:
k(x+3)(x6)=0 or
k(x²3x18)=0 ...(1)
where k is a constant to be determined.
Since the yintercept is negative, k>0.
If ∠ABC is a right angle, then a circle with AB as diameter intersects the yaxis at intercepts (0,p) and (0,p) where p<0.
The centre of circle ABC is at ((63)/2,0), or (1.5,0), and the radius is (6(3))/2 = 4.5.
The equation of circle ABC is therefore:
C1 : (x1.5)^2+y^2=4.5^2, or
C1: (x1.5)^2 + y^2 = 20.25
To find intersection with the yaxis, 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 yintercepts), 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.