2004 AMC 12B Problems/Problem 21

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 $a+b$?

$\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ \sqrt{10} \qquad\mathrm{(C)}\ \frac 72 \qquad\mathrm{(D)}\ \frac 92 \qquad\mathrm{(E)}\ 2\sqrt{14}$

Solution

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, \[(20m+11)^2 - 4\cdot 40 \cdot (3m^2 + m + 2) = -80m^2 + 280m - 199\] 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions

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