AMC12 2004 B

AMC12 2004 B · Q21

AMC12 2004 B · Q21. It mainly tests Quadratic equations, Circle theorems.

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 $a+b$?

方程 $2x^2 + xy + 3y^2 - 11x - 20y + 40 = 0$ 的图像是 $xy$ 平面第一象限内的一个椭圆。设 $a$ 和 $b$ 分别为椭圆上所有点 $(x,y)$ 的 $\frac yx$ 的最大值与最小值。求 $a+b$ 的值。

(A) 3 3

(B) \sqrt{10} \sqrt{10}

(D) \frac{9}{2} \frac{9}{2}

(E) 2\sqrt{14} 2\sqrt{14}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Error creating thumbnail: Unable to save thumbnail to destination $\frac yx$ 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 $\frac{280}{80} = \frac{7}{2} \Rightarrow \mathrm{(C)}$.

$\frac yx$ 表示过原点的一条直线的斜率。由于直线 $y=mx$ 与椭圆的交点个数可能为 $0,1,$ 或 $2$ 个，因此当 $y=mx$ 与椭圆相切（只有一个交点）时，$\frac yx$ 取得最小值与最大值。代入得 \[2x^2 + x(mx) + 3(mx)^2 - 11x - 20(mx) + 40 = 0\] 按 $x$ 的次数整理： \[(3m^2 + m + 2)x^2 - (20m + 11)x + 40 = 0\] 由于直线 $y=mx$ 与椭圆相切，判别式 \[(20m+11)^2 - 4\cdot 40 \cdot (3m^2 + m + 2) = -80m^2 + 280m - 199\] 应等于 $0$。所求 $a+b$ 为上述二次方程两根之和。由韦达定理， $\frac{280}{80} = \frac{7}{2} \Rightarrow \mathrm{(C)}$.

Topics