1. Determine the value of the parameter $n$ such that for every point $(x, y)$ on the graph of the function $y=-x+n$, the following equations hold: $y=\frac{5}{x}$ and $y^{2}=6-x^{2}$. Then, calculate the area of the region in the plane formed by the coordinate axes and the graph of the function.
Solution. From the condition $y^{2}=6-x^{2}$, we have $x^{2}+y^{2}=6$, and from $y=\frac{5}{x}$, we get $x y=5, x \neq 0$. By squaring $y=-x+n$, we obtain

$$
x^{2}+y^{2}+2 x y=n^{2}
$$

Accordingly, $n^{2}=6+10$, and thus $n= \pm 4$. Therefore, we obtain two functions $y=-x+4$ and $y=-x-4$. For each of them, the part of the plane they form with the coordinate axes is a right triangle with the length of the legs

![](https://cdn.mathpix.com/cropped/2024_06_05_ee54e896908301490c8ag-7.jpg?height=644&width=565&top_left_y=853&top_left_x=914)
equal to 4, so the desired area is $P=\frac{4 \cdot 4}{2}=8$.