2004 AMC 12B Problem 21

Attempt Problem 21 of the 2004 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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21.

The graph of 2x2+xy+3y22x^2 + xy + 3y^2 11x20y+40=0- 11x - 20y + 40 = 0 is an ellipse in the first quadrant of the xyxy-plane. Let aa and bb be the maximum and minimum values of yx\dfrac{y}{x} over all points (x,y)(x, y) on the ellipse. What is the value of a+b?a + b?

33

10\sqrt{10}

72\dfrac{7}{2}

92\dfrac{9}{2}

2142\sqrt{14}

Answer: C
Concepts:ellipsetangent lineVieta’s Formulas
Difficulty rating: 2080
Solution:

The slopes aa and bb are the values of mm for which y=mxy = mx meets the ellipse in exactly one point. Substituting gives (3m2+m+2)x2(20m+11)x+40=0. \begin{aligned} &(3m^2 + m + 2)x^2 \\ &\quad {}- (20m + 11)x + 40 = 0. \end{aligned} Setting its discriminant to zero yields 80m2+280m199=0.-80m^2 + 280m - 199 = 0. By Vieta's formulas, a+b=28080=72.a + b = \dfrac{280}{80} = \dfrac{7}{2}.

Thus, the correct answer is C.

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