1. Let the pair (1,1) be a solution of the equation $\frac{x}{a}+\frac{y}{b}=1$, where $a$ and $b$ are non-zero real parameters. Determine $a$ and $b$ if $a^{2}+b^{2}=\frac{45}{4}$.
Solution. From $\frac{1}{a}+\frac{1}{b}=1$ it follows that $a+b=a b$. Therefore,

$$
\frac{45}{4}=a^{2}+b^{2}=(a+b)^{2}-2 a b=(a b)^{2}-2 a b
$$

The solutions to the quadratic equation $(a b)^{2}-2 a b-\frac{45}{4}=0$ are $1 \pm \frac{7}{2}$, so $a b=\frac{9}{2}$ or $a b=-\frac{5}{2}$. Accordingly, $a+b=a b=\frac{9}{2}$ or $a+b=a b=-\frac{5}{2}$. Therefore, $a$ and $b$ are solutions to the quadratic equations $z^{2}-\frac{9}{2} z+\frac{9}{2}=0$ or $z^{2}+\frac{5}{2} z-\frac{5}{2}=0$. The solutions to the first equation are $z_{1,2}=\frac{9 \pm 3}{4}$, and to the second equation are $z_{3,4}=\frac{-5 \pm \sqrt{65}}{4}$.

Finally, $(a, b)=\left(3, \frac{3}{2}\right),\left(\frac{3}{2}, 3\right),\left(\frac{-5+\sqrt{65}}{4}, \frac{-5-\sqrt{65}}{4}\right),\left(\frac{-5-\sqrt{65}}{4}, \frac{-5+\sqrt{65}}{4}\right)$.