123. Find the coordinates of the foci, the lengths of the axes, and the eccentricity of the ellipse given by the equation $2 x^{2}+y^{2}=32$.
Solution. We will bring the equation of the ellipse to the canonical form (2). For this, we divide all its terms by 32:

$$
\frac{2 x^{2}}{32}+\frac{y^{2}}{32}=\frac{32}{32} ; \frac{x^{2}}{16}+\frac{y^{2}}{32}=1
$$

We see that $a^{2}=16, b^{2}=32$, from which $a=4, b=4 \sqrt{2}$.

Since $b>a$, the foci of the ellipse are located on the ordinate axis. They have coordinates $F_{1}(0 ; c)$ and $F_{2}(0 ;-c)$, where $c$ is determined from the relation $b^{2}-a^{2}=c^{2}$. Substituting the values of $a$ and $b$ into it, we get $32-16=c^{2} ; c^{2}=16 ; c=4$. Thus, the foci of the ellipse are the points $F_{1}(0 ; 4) ; F_{2}(0 ;-4)$.

Next, we find: the major axis of the ellipse $2 b=8 \sqrt{2}$; the minor axis $2 a=8$; the eccentricity $\varepsilon=\frac{2 c}{2 b}=\frac{8}{8 \sqrt{2}}=\frac{\sqrt{2}}{2}=0.705$.