Task B-4.6. Calculate the area of the quadrilateral bounded by the common tangents of the curves $x^{2}+2 y^{2}=2$ and $4 x^{2}-9 y^{2}=36$.
## Solution.

Image.

(2 points)

The curve $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ is an ellipse and it holds that $a^{2}=2, b^{2}=1$. Using the tangency condition $a^{2} k^{2}+b^{2}=l^{2}$ for a line and an ellipse, we get

$$
2 k^{2}+1=l^{2} .(*)
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-24.jpg?height=646&width=1100&top_left_y=200&top_left_x=488)

The curve $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ is a hyperbola and it holds that $a^{2}=9, b^{2}=4$. Using the tangency condition $a^{2} k^{2}-b^{2}=l^{2}$ for a line and a hyperbola, we get

$$
9 k^{2}-4=l^{2} \cdot(* *)
$$

From the system of equations $(*)$ and $(* *)$, we obtain the solutions:

$$
k_{1,2}= \pm \sqrt{\frac{5}{7}} \text{ and } l_{1,2}= \pm \sqrt{\frac{17}{7}}
$$

The equations of the common tangents are

$$
\begin{aligned}
t_{1,2,3,4} \ldots y & = \pm \sqrt{\frac{5}{7}} x \pm \sqrt{\frac{17}{7}} \\
y & = \pm \sqrt{\frac{5}{7}} x \mp \sqrt{\frac{17}{7}}
\end{aligned}
$$

Let's calculate their intercepts on the coordinate axes.

On the $x$-axis for $y=0$, we get

$$
\begin{gathered}
x_{1,2}=\mp \sqrt{\frac{17}{5}} \\
d_{1}=x_{2}-x_{1}=2 \sqrt{\frac{17}{5}}
\end{gathered}
$$

On the $y$-axis for $x=0$, we get

$$
\begin{gathered}
y_{1,2}= \pm \sqrt{\frac{17}{7}} \\
d_{1}=y_{2}-y_{1}=2 \sqrt{\frac{17}{7}}
\end{gathered}
$$

Then the desired area is

$$
P=\frac{1}{2} d_{1} d_{2}=\frac{1}{2} \cdot 2 \sqrt{\frac{17}{5}} \cdot \sqrt{\frac{17}{7}}=\frac{34 \sqrt{35}}{35}
$$