2. Find the solution of the system

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=a^{2}, a>0 \\
z^{2}+t^{2}=b^{2}, \quad b>0 \\
\quad x t+y z=a b
\end{array}\right.
$$

for which the value of $x+z$ is maximized.

- The left side of the last equation resembles the dot product in coordinate form. Let's use this. Consider it as the dot product of vectors $\vec{n}\{x ; y\}$ and $\vec{m}\{t ; z\}$. Then

$$
\begin{gathered}
|\vec{n}|=\sqrt{x^{2}+y^{2}}=a, \quad|\vec{m}|=\sqrt{z^{2}+t^{2}}=b \\
\vec{n} \cdot \vec{m}=x t+y z=a b
\end{gathered}
$$

On the other hand, by the definition of the dot product,

$$
\vec{n} \cdot \vec{m}=a b \cos \alpha ; \alpha-\text { angle between the vectors; }
$$

$\vec{n} \cdot \vec{m}=a b \Rightarrow \cos \alpha=1$; the vectors are collinear, their

coordinates are proportional,

$$
\frac{x}{t}=\frac{y}{z}=\frac{a}{b} ; z=\frac{b}{a} y .
$$

Then

$$
x+z=\sqrt{a^{2}-y^{2}}+\frac{b}{a} y
$$

The function $f(y)=\sqrt{a^{2}-y^{2}}+\frac{b}{a} y$ reaches its maximum $\sqrt{a^{2}+b^{2}}$ at $y=$ $\frac{a b}{\sqrt{a^{2}+b^{2}}} \cdot$ From this,

$$
\begin{gathered}
z=\frac{b}{a} y=\frac{b^{2}}{\sqrt{a^{2}+b^{2}}} \\
t^{2}=b^{2}-z^{2}=b^{2}-\frac{b^{4}}{a^{2}+b^{2}}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \Rightarrow t=\frac{a b}{\sqrt{a^{2}+b^{2}}} \\
x^{2}=a^{2}-y^{2}=a^{2}-\frac{a^{2} b^{2}}{a^{2}+b^{2}}=\frac{a^{4}}{a^{2}+b^{2}} \Rightarrow x=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}
\end{gathered}
$$

Thus, the solution that satisfies the condition of the problem is found:

$$
x=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ; y=\frac{a b}{\sqrt{a^{2}+b^{2}}} ; z=\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}, t=\frac{a b}{\sqrt{a^{2}+b^{2}}}
$$

We will prove that in any solution of the system, the numbers $x$ and $t$ have the same sign, as well as the numbers $y$ and $z$. Both products $x t$ and $y z$ cannot be negative at the same time. Suppose one is positive and the other is negative; for example, $x t>0, y z<0$. Since $|x| \leq a$ and $|t| \leq b$, then $x t \leq a b \Rightarrow x t+y z<a b$. Contradiction. Taking this into account, all other solutions can be obtained by selecting the signs plus or minus

|  | $x$ | $y$ | $z$ | $t$ |
| :---: | :---: | :---: | :---: | :---: |
| 1 | + | + | + | + |
| 2 | + | - | - | + |
| 3 | - | + | + | - |
| 4 | - | - | - | - |

- Another solution. The form of the first equation of the system $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$ reminds us of the Pythagorean theorem and the fundamental trigonometric identity. The substitution $\frac{x}{a}=\cos \varphi, \frac{y}{a}=\sin \varphi, z=b \cos \psi, t=b \sin \psi$ is logical. From the last equation of the system, we have $\cos \varphi \sin \psi+\sin \varphi \cos \psi=1$;

$$
\begin{gathered}
\sin (\varphi+\psi)=1 ; \quad \varphi+\psi=\frac{\pi}{2}+2 \pi k, k \in Z ; \quad \psi=\frac{\pi}{2}+2 \pi k-\varphi . \\
x+z=a \cos \varphi+b \cos \psi=a \cos \varphi+b \sin \varphi=\sqrt{a^{2}+b^{2}}(\cos (\varphi+\omega)) ; \\
\sin \omega=\frac{b}{\sqrt{a^{2}+b^{2}}} ; \quad \cos \omega=\frac{a}{\sqrt{a^{2}+b^{2}}} ; \quad \operatorname{tg} \omega=\frac{b}{a}
\end{gathered}
$$

We performed the transformation using the method of introducing an auxiliary angle.

The maximum value $\sqrt{a^{2}+b^{2}}$ is achieved when $\cos (\varphi+\omega)=1$;

$\varphi+\omega=2 \pi k, k \in Z ; \varphi=2 \pi k-\omega ; \cos \varphi=\cos \omega=\frac{a}{\sqrt{a^{2}+b^{2}}} ; x=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$.
# Answer:

$\left(\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ; \frac{a b}{\sqrt{a^{2}+b^{2}}} ; \frac{b^{2}}{\sqrt{a^{2}+b^{2}}}, \frac{a b}{\sqrt{a^{2}+b^{2}}}\right),\left(\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ;-\frac{a b}{\sqrt{a^{2}+b^{2}}} ;-\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}, \frac{a b}{\sqrt{a^{2}+b^{2}}}\right)$, $\left(-\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ; \frac{a b}{\sqrt{a^{2}+b^{2}}} ; \frac{b^{2}}{\sqrt{a^{2}+b^{2}}},-\frac{a b}{\sqrt{a^{2}+b^{2}}}\right),\left(-\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} ;-\frac{a b}{\sqrt{a^{2}+b^{2}}} ;-\frac{b^{2}}{\sqrt{a^{2}+b^{2}}},-\frac{a b}{\sqrt{a^{2}+b^{2}}}\right)$