3A. Determine the tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ that forms a triangle with the coordinate axes having the smallest hypotenuse.
Solution. Let the line $y=k x+n$ be a tangent to the given ellipse. According to this, the system

$$
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\
y=k x+n
\end{array}\right.
$$

has a unique solution. Substituting $y$ from the second equation into the first, we get a quadratic equation in $y$ and using the fact that its discriminant is equal to 0, we obtain

![](https://cdn.mathpix.com/cropped/2024_06_05_46a9829282eea0e99a44g-09.jpg?height=404&width=618&top_left_y=1516&top_left_x=842)
the condition for the line to be tangent to the ellipse $a^{2} k^{2}+b^{2}=n^{2}$. If $k=0$ or $n=0$, the tangent is parallel to one of the coordinate axes, so it does not form a triangle. Therefore, we can assume that $k \neq 0$ and $n \neq 0$. Furthermore, if we write the equation of the tangent in segment form, we get $\frac{x}{-\frac{n}{k}}+\frac{y}{n}=1$, so the length of the segment between the coordinate axes is $d=\sqrt{\frac{n^{2}}{k^{2}}+n^{2}}$. According to this,

$$
d^{2}=\frac{n^{2}}{k^{2}}+n^{2}=\frac{a^{2} k^{2}+b^{2}}{k^{2}}+a^{2} k^{2}+b^{2}=a^{2} k^{2}+\frac{b^{2}}{k^{2}}+a^{2}+b^{2}=\left(a k-\frac{b}{k}\right)^{2}+(a+b)^{2}
$$

The length of the segment will be the smallest if $a k-\frac{b}{k}=0$. From this, $k= \pm \sqrt{\frac{b}{a}}$ and $n= \pm \sqrt{a b+b^{2}} \quad(a, b>0)$. Therefore, there are four lines with the desired property, and they are

$$
\begin{aligned}
& y=\sqrt{\frac{b}{a}} x+\sqrt{a b+b^{2}}, y=\sqrt{\frac{b}{a}} x-\sqrt{a b+b^{2}} \\
& y=-\sqrt{\frac{b}{a}} x+\sqrt{a b+b^{2}}, y=-\sqrt{\frac{b}{a}} x-\sqrt{a b+b^{2}}
\end{aligned}
$$