131. Find the coordinates of the foci, the lengths of the axes, the eccentricity, and the equations of the asymptotes of the hyperbola given by the equation $16 x^{2}-$ $-25 y^{2}=400$.
Solution. We will bring the equation to the canonical form (3). For this, we divide all its terms by 400:

$$
\frac{16 x^{2}}{400}-\frac{25 y^{2}}{400}=\frac{400}{400} ; \quad \frac{x^{2}}{25}-\frac{y^{2}}{16}=1
$$

From this equation, it is clear that the foci of the hyperbola are located on the x-axis, and we can write $a^{2}=25, b^{2}=16$, i.e., $a=5, b=4$. Knowing $a$ and $b$, we find $c: a^{2}+b^{2}=c^{2} ; c^{2}=25+16=41 ; c=\sqrt{41}$.

Thus, the foci of the hyperbola are the points $F_{1}(-\sqrt{41} ; 0) ; F_{2}(\sqrt{41} ; 0)$; the real axis of the hyperbola $2 a=10$; the imaginary axis $2 b=8$; the eccentricity $\varepsilon=2 \sqrt{41} / 10=\sqrt{41} / 5$; the equations of the asymptotes $y= \pm(4 / 5) x$.