3. Determine the tangent to the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ whose segment between the axes of the ellipse is the smallest.
Solution. Since the ellipse is symmetric with respect to the coordinate axes, it is sufficient to determine the tangent that touches the ellipse at the point $(x_{0}, y_{0})$ which lies in the first quadrant. The equation of the tangent is

$$
\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1
$$

the segments it intercepts on the coordinate axes are equal to $\frac{a^{2}}{x_{0}}$ and $\frac{b^{2}}{y_{0}}$, and the square of the length of the segment of the tangent between the axes is equal to

$$
\begin{aligned}
d^{2} & =\frac{a^{4}}{x_{0}^{2}}+\frac{b^{4}}{y_{0}^{2}}=\frac{a^{4}}{x_{0}^{2}}+\frac{a^{2} b^{2}}{a^{2}-x_{0}^{2}} \\
& =a^{2}+b^{2}+\frac{a^{2}\left(a^{2}-x_{0}^{2}\right)}{x_{0}^{2}}+\frac{b^{2} x_{0}^{2}}{a^{2}-x_{0}^{2}} \\
& \geq a^{2}+b^{2}+2 \sqrt{\frac{a^{2}\left(a^{2}-x_{0}^{2}\right)}{x_{0}^{2}} \cdot \frac{b^{2} x_{0}^{2}}{a^{2}-x_{0}^{2}}} \\
& =(a+b)^{2}
\end{aligned}
$$

where the equality sign holds if and only if $\frac{y_{0}}{x_{0}}=\left(\frac{b}{a}\right)^{\frac{3}{2}}$. Therefore, the desired tangent (in the first quadrant) is the one whose point of tangency $(x_{0}, y_{0})$ satisfies the given condition. Its equation is

$$
y=-\sqrt{\frac{b}{a}} x+\sqrt{a b+b^{2}}
$$