Example 4. From a point $\mathrm{P}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ outside the parabola $\mathrm{y}^{2}=4 \mathrm{x}$, two tangents $P A$ and $P B$ are drawn, with $A$ and $B$ being the points of tangency.
(1) Express the area of $\triangle P A B$ in terms of $x_{0}, y_{0}$;
(2) If $\mathrm{P}\left(x_{0}, y_{0}\right)$ moves on the circle $(x+4)^{2}+y^{2}=9$, at what position of $\mathrm{P}$ is the area of $\triangle P A B$ maximized, and what is the maximum area?
(1) As shown in Figure 5, let the points of tangency be \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then the equation of the chord of contact \( AB \) is:
$$
y_{0} y=2\left(x+x_{0}\right),
$$

which simplifies to \( 2 x - y_{0} y + 2 x_{0} = 0 \).
Draw \( PC \) perpendicular to the chord \( AB \) intersecting \( AB \) at \( C \),

then \( |PC| = \frac{\left|4 x_{0} - y_{0}^2\right|}{\sqrt{y_{0}^2 + 4}} \). From the diagram, we know that
$$
|AB| = \frac{\left|y_{1} - y_{2}\right|}{\sin \alpha} \quad \text{(or } |AB| = \frac{\left|y_{1} - y_{2}\right|}{\sin (\pi - \alpha)} \text{)}.
$$
First, find \( \sin \alpha \). From
$$
k_{AB} = \tan \alpha = \frac{2}{y_{0}},
$$
we have
$$
\frac{\sin \alpha}{\cos \alpha} = \frac{2}{y_{0}},
$$
which gives
$$
\sin \alpha = \frac{2}{\sqrt{y_{0}^2 + 4}}.
$$

Next, find \( \left|y_{1} - y_{2}\right| \).
Solve the system of equations:
$$
\left\{\begin{array}{l}
y^2 = 4x, \\
2x - y_{0} y + 2x_{0} = 0
\end{array}\right.
$$
From (2), we get \( x = \frac{y_{0} y - 2x_{0}}{2} \). Substituting into (1) and simplifying,

we obtain: \( y^2 - 2y_{0} y + 4x_{0} = 0 \).
By Vieta's formulas: \( y_{1} + y_{2} = 2y_{0}, y_{1} y_{2} = 4x_{0} \).
Thus,
$$
\left|y_{1} - y_{2}\right| = \sqrt{(y_{1} + y_{2})^2 - 4y_{1} y_{2}} = 2 \sqrt{y_{0}^2 - 4x_{0}}.
$$
Therefore,
$$
|AB| = \sqrt{y_{0}^2 + 4} \cdot \sqrt{y_{0}^2 - 4x_{0}}.
$$
Hence,
$$
S_{\triangle PAB} = \frac{1}{2} |PC| \cdot |AB| = \frac{1}{2} \left(y_{0}^2 - 4x_{0}\right)^{\frac{3}{2}}.
$$
(2) Since \( P(x_{0}, y_{0}) \) lies on \( (x + 4)^2 + y^2 = 9 \),
$$
\therefore y_{0}^2 = 9 - (x_{0} + 4)^2.
$$

Substituting this into the above expression, we get:
$$
\begin{aligned}
S_{\triangle PAB} & = \frac{1}{2} \left[9 - (x_{0} + 4)^2 - 4x_{0}\right]^{\frac{3}{2}} \\
& = \frac{1}{2} \left[-(x_{0} + 6)^2 + 29\right]^{\frac{3}{2}}.
\end{aligned}
$$

When \( x_{0} = -6 \), \( S_{\triangle PAB} \) reaches its maximum value \( \frac{29 \sqrt{29}}{2} \).
Substituting \( x_{0} = -6 \) into \( y_{0}^2 = 9 - (x_{0} + 4)^2 \), we get: \( y_{0} = \pm \sqrt{5} \).
Thus, when \( P \) is at \( (-6, -\sqrt{5}) \) or \( (-6, \sqrt{5}) \), the area of \( \triangle PAB \) is maximized, with the maximum value being \( \frac{29 \sqrt{29}}{2} \).