## Task A-4.2.

The set of all points $(x, y)$ for which $y^{2}+2 x y+40|x|=400$ divides the plane into several parts, of which only one is bounded. Determine the area of this bounded part of the plane.
## Solution.

Notice that $y^{2}+2 x y+40|x|=400$ is equivalent to

$$
y^{2}+2 x y+x^{2}=x^{2}-40|x|+400
$$

or

$$
(y+x)^{2}=(|x|-20)^{2}
$$

thus $y+x= \pm(|x|-20)$.

The set of all given points are actually four rays determined by the following equations:

$$
\begin{array}{lll}
y=-x+(-x-20)=-2 x-20 & \text { for } x \leqslant 0, & 1 \text { point } \\
y=-x+(x-20)=-20 & \text { for } x \geqslant 0, & 1 \text { point } \\
y=-x-(-x-20)=20 & \text { for } x \leqslant 0, & 1 \text { point } \\
y=-x-(x-20)=-2 x+20 & \text { for } x \geqslant 0, & 1 \text { point }
\end{array}
$$

The only bounded region they determine is a parallelogram with vertices $A_{1}(0,-20)$, $A_{2}(20,-20), A_{3}(0,20)$, and $A_{4}(-20,20)$.

![](https://cdn.mathpix.com/cropped/2024_05_30_14eaa2cf500c23a22c24g-18.jpg?height=702&width=694&top_left_y=1365&top_left_x=607)

Its area is $\left|A_{1} A_{2}\right| \cdot\left|A_{1} A_{3}\right|=20 \cdot 40=800$.