AMC12 2004 12B, Problem 21

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 $a+b$?

Solution: $\frac yx$ 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 $y=mx$ is tangent to the ellipse, 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} $. $\fbox{\frac 72}$.

Answer: \frac 72