2004 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2004 AMC 12B, 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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Concepts:ellipse tangent line Vieta’s Formulas

Difficulty rating: 2080

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 $\dfrac{y}{x}$ over all points $(x, y)$ on the ellipse. What is the value of $a + b?$

$3$

$\sqrt{10}$

$\dfrac{7}{2}$

$\dfrac{9}{2}$

$2\sqrt{14}$

Solution:

The slopes $a$ and $b$ are the values of $m$ for which $y = mx$ meets the ellipse in exactly one point. Substituting gives $(3m^2 + m + 2)x^2 - (20m + 11)x + 40 = 0.$ Setting its discriminant to zero yields $-80m^2 + 280m - 199 = 0.$ By Vieta's formulas, $a + b = \dfrac{280}{80} = \dfrac{7}{2}.$

Thus, the correct answer is C.

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