2004 AMC 12B Problem 21

Question

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}

The extreme values of

\dfrac{y}{x}

come from lines

y = mx

tangent to the ellipse

Substitute

y = mx

and set the discriminant of the resulting quadratic in

x

to zero

The slopes

a

and

b

are the values of

m

for which

y = mx

meets the ellipse in exactly one point. Substituting gives

\begin{aligned} &(3m^2 + m + 2)x^2 \\ &\quad {}- (20m + 11)x + 40 = 0. \end{aligned}

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.

Problem 21 in other years