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

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

== Solution ==
[[Image:2004_12B_AMC-21.png]]

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

== See also ==
{{AMC12 box|year=2004|ab=B|num-b=20|num-a=22}}

[[Category:Introductory Algebra Problems]]
[[Category:Introductory Geometry Problems]]