Task B-4.4. The vertices of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, on the x-axis, are also the foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, and the vertices of this hyperbola are also the foci of the given ellipse. Determine the tangent of the angle that the asymptotes of the hyperbola enclose.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-17.jpg?height=626&width=1022&top_left_y=1752&top_left_x=519)

Since for the ellipse $e^{2}=a^{2}-b^{2}=25-9=16$, the foci of the ellipse are $(-4,0)$ and $(4,0)$.

(1 point)

Then the real semi-axis of the hyperbola is $a_{h}=4$.

The vertices of the ellipse on the x-axis are the points $(-5,0)$ and $(5,0)$. These are also the foci of the hyperbola, so the linear eccentricity of the hyperbola is $e_{h}=5$.

(1 point)

Then $b_{h}=\sqrt{25-16}=3$.

The asymptotes of the hyperbola are the lines $y= \pm \frac{3}{4} x$, and then the tangent of the angle between them is

$$
\operatorname{tg} \varphi=\left|\frac{\frac{3}{4}+\frac{3}{4}}{1-\frac{3}{4} \cdot \frac{3}{4}}\right|=\frac{24}{7}
$$