4. Determine the tangent of the angle formed by the common tangents of the curves $x^{2}+2 y^{2}=2$ and $y^{2}=4 x$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d793ff9f9da41a4dd093g-17.jpg?height=896&width=1145&top_left_y=1525&top_left_x=544)

The condition for the line $y=k x+l$ to be tangent to an ellipse with semi-axes $a$ and $b$ is $a^{2} k^{2}+b^{2}=l^{2}$, while the condition for the line $y=k x+l$ to be tangent to the parabola $y^{2}=2 p x$ is given by $p=2 k l$.

Thus, we obtain a system of equations with unknowns $k$ and $l$:

$$
2 k^{2}+1=l^{2}, \quad 2=2 k l
$$

from which we get a biquadratic equation in $k$:

$$
2 k^{4}+k^{2}-1=0
$$

whose real solutions are $k_{1}=\frac{\sqrt{2}}{2}=\operatorname{tg} \varphi_{1}$ and $k_{2}=-\frac{\sqrt{2}}{2}=\operatorname{tg} \varphi_{2}$, from which it follows that

$$
\operatorname{tg} \varphi=\frac{\operatorname{tg} \varphi_{1}-\operatorname{tg} \varphi_{2}}{1+\operatorname{tg} \varphi_{1} \operatorname{tg} \varphi_{2}}=\frac{2 k_{1}}{1-k_{1}^{2}}=\frac{2 \frac{\sqrt{2}}{2}}{1-\left(\frac{\sqrt{2}}{2}\right)^{2}}=2 \sqrt{2}
$$