Question

一个四边形的内部由图像和围成，其中 $a$ 是正实数。对所有 $a > 0$ ，这个区域的面积用 $a$ 表示是多少？ $(x+ay)^2=4a^2$ $(ax-y)^2=a^2$ ，

The interior of a quadrilateral is bounded by the graphs of $(x+ay)^2=4a^2$ and $(ax-y)^2=a^2,$ where $a$ is a positive real number. What is the area of this region in terms of $a,$ valid for all $a > 0?$

$\dfrac{8a^2}{(a+1)^2}$

$\dfrac{4a}{a+1}$

$\dfrac{8a}{a+1}$

$\dfrac{8a^2}{a^2+1}$

$\dfrac{8a}{a^2+1}$

答案：D

知识点：平行线距离公式矩形

难度评级：1720

解答：

注意，每个方程都会给出两条平行直线。

给出两条直线和这两条直线的斜率都是 $-\dfrac{1}{a}$ 。 $(x + ay)^2 = 4a^2$ $x + ay - 2a = 0$ $x + ay + 2a = 0.$

类似地，给出直线和这些直线的斜率为 $a$ 。 $(ax-y)^2 = a^2$ $ax - y - a = 0$ $ax - y + a = 0.$

两组直线互相垂直，因此围成一个矩形。

回忆两条平行线之间的距离 $d$ 为 $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ $d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.$

用这个公式，第一组平行线之间的距离为类似地，第二组平行线之间的距离为 $\dfrac{4a}{\sqrt{a^2 + 1}}.$ $\dfrac{2a}{\sqrt{a^2 + 1}}.$

这两个距离就是矩形的边长，相乘得面积 $\dfrac{8a^2}{a^2 + 1}.$

所以正确答案是 D。

Note that each of the equations yields two parallel lines.

$(x + ay)^2 = 4a^2$ results in the two lines $x + ay - 2a = 0$ and $x + ay + 2a = 0.$ Both of these lines have a slope of $-\dfrac{1}{a}.$

Similarly, $(ax-y)^2 = a^2$ results in the lines $ax - y - a = 0$ and $ax - y + a = 0.$ These lines have slope $a.$

Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle.

Recall that the formula for the distance $d$ between two parallel lines $\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$ is $d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.$

Using this formula, we get that the distance between the first pair of lines is $\dfrac{4a}{\sqrt{a^2 + 1}}.$ Similarly, the distance between the second pair of lines is $\dfrac{2a}{\sqrt{a^2 + 1}}.$

These are the side lengths of the rectangle. Multiplying yields the area $\dfrac{8a^2}{a^2 + 1}.$

Thus, D is the correct answer.

2021 AMC 10A Spring 第 24 题

答案：D

难度评级：1720

解答：

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