Difficult Quadratics & the Parabola question (1 Viewer)

[c_james]

Any help with this one would be much appreciated:

The parabola y = ax² + bx + c (where c does not equal 0) meets the x axis at A(α, 0) and B(β, 0) and the y-axis at C. If AC and BC are perpendicular, prove that AC = -1.

 

[CM_Tutor]

As A is at (α, 0), B at (β, 0), and C at (0, c):

m_AC = (c - 0) / (0 - α) = -c / α
Similarly, m_BC = -c / β

Now AC _|_ BC, so m_AC * m_BC = -1
So, (-c / α) * (-c / β) = -1
So, αβ = -c²

Now, since A and B are the x-intercepts of y = ax² + bx + c, it follows that α and β are the roots of ax² + bx + c = 0
So, αβ = c / a

So, -c² = c / a, as both equal αβ
So, ac = -1, as required, as c ≠ 0

 

[c_james]

Thanks a bunch, its not too difficult on closer inspection.