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Difficult Quadratics & the Parabola question (1 Viewer)

c_james

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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

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

m<sub>AC</sub> = (c - 0) / (0 - &alpha;) = -c / &alpha;
Similarly, m<sub>BC</sub> = -c / &beta;

Now AC _|_ BC, so m<sub>AC</sub> * m<sub>BC</sub> = -1
So, (-c / &alpha;) * (-c / &beta;) = -1
So, &alpha;&beta; = -c<sup>2</sup>

Now, since A and B are the x-intercepts of y = ax<sup>2</sup> + bx + c, it follows that &alpha; and &beta; are the roots of ax<sup>2</sup> + bx + c = 0
So, &alpha;&beta; = c / a

So, -c<sup>2</sup> = c / a, as both equal &alpha;&beta;
So, ac = -1, as required, as c &ne; 0
 

c_james

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

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