Example 4. Given the hyperbola $x y=1$, find the tangent line that is intercepted by the hyperbola $4 x y=3$ to form the shortest chord. Determine the equation of the tangent line, and under what conditions the chord is the shortest, as well as the length of the shortest chord.

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Let the equation of the tangent line be: $y=kx+b$. Substituting into $xy=1$ yields $kx^{2}+bx-1=0$.
$\because$ it is tangent,
$\therefore \triangle=b^{2}+4k=0, k=-\frac{b^{2}}{4}$. Therefore, the equation of the tangent line is $y=-\frac{b^{2}}{4}x+b$.

Substituting $y=kx+b$ into $4xy=3$, we get $b^{2}x^{2}-4bx+3=0$. Let the two intersection points of the line $y=-\frac{b^{2}}{4}x+b$ with the curve $4xy=3$ be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. By the slope formula,
$$
\begin{array}{l}
k=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\frac{b^{2}}{4}, \text{ so } (y_{1}-y_{2})^{2} \\
=\frac{b^{4}}{16}(x_{1}-x_{2})^{2}. \\
\therefore |AB|=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}} \\
=\sqrt{(x_{1}-x_{2})^{2}+\frac{b^{4}}{16}(x_{1}-x_{2})^{2}} \\
=\sqrt{\left(1+\frac{b^{4}}{16}\right)(x_{1}-x_{2})^{2}} \\
=\sqrt{\left(1+\frac{b^{4}}{16}\right)\left[(x_{1}+x_{2})^{2}-4x_{1}x_{2}\right]} \\
=\sqrt{\frac{16+b^{4}}{16}\left[\frac{16}{b^{2}}-\frac{12}{b^{2}}\right]} \\
=\sqrt{\frac{4}{b^{2}}+\frac{b^{2}}{4}} \geqslant \sqrt{2}. \\
\end{array}
$$
$\therefore$ When $\frac{4}{b^{2}}=\frac{b^{2}}{4}$, i.e., $b= \pm 2$, $|AB|$ has the minimum value $\sqrt{2}$. At this time, $k=-1$, and the chord length $|AB|=\sqrt{2}$. The equation of the tangent line is $y=-x \pm 2$.
(2) Using the parametric equation of the line, convert to Vieta's formulas.

Let the parametric equation of the line passing through $p_{0}\left(x_{0}, y_{0}\right)$ be $\left\{\begin{array}{l}x=x_{0}+t \cos \alpha, \\ y=y_{0}+t \sin \alpha .\end{array}(t\right.$ is the parameter, $0 \leqslant \alpha<\pi)$. By the geometric meaning of parameter $t$, we have (1) $t_{\text {mid}}=\frac{t_{1}+t_{2}}{2}$;
(2) $|AB|=\left|t_{1}-t_{2}\right|=\sqrt{\left(t_{1}+t_{2}\right)^{2}-4t_{1}t_{2}}$; (3) $\left|P_{0}A\right| \cdot \left|P_{0}B\right|=\left|t_{1}t_{2}\right|$. Here, $t_{1}$, $t_{2}$, and $t_{\text{mid}}$ are the positions of $A$, $B$, and $P$ relative to $P_{0}$, respectively. This can be used to derive the required results.