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2004 AMC 12B Problems/Problem 21

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Problem

The graph of 2x^2 + xy + 3y^2 - 11x - 20y + 40 = 0 is an ellipse in the first quadrant of the xy-plane. Let a and b be the maximum and minimum values of \frac yx over all points (x,y) on the ellipse. What is the value of a+b?

\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ \sqrt{10}\qquad\mathrm{(C)}\ \frac 72\qquad\mathrm{(D)}\ \frac 92\qquad\mathrm{(E)}\ 2\sqr...

Solution

Image:2004_12B_AMC-21.png

\frac yx represents the slope of a line passing through the origin. It follows that since a line y = mx intersects the ellipse at either 0, 1, or 2 points, the minimum and maximum are given when the line y = mx is a tangent, with only one point of intersection. Substituting, 2x^2 + x(mx) + 3(mx)^2 - 11x - 20(mx) + 40 = 0 Rearranging by the degree of x, (3m^2 + m + 2)x^2 - (20m + 11)x + 40 = 0 Since the line y=mx, we want the discriminant, (20m+11)^2 - 4\cdot 40 \cdot (3m^2 + m + 2) = -80m^2 + 280m - 199 = 0 to be equal to 0. We want a+b, which is the sum of the roots of the above quadratic. By Vieta’s formulas, that is \frac{280}{80} = \frac{7}{2} \Rightarrow \mathrm{(C)}.

See also

2004 AMC 12B (ProblemsResources)
Preceded by
Problem 20
Followed by
Problem 22
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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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