Example 3. Find the equation of the line passing through $\mathrm{A}$ $(-6,3)$ and tangent to the parabola $2 y^{2}=9 x$.

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Let the required tangent line equation be
$$
\left\{\begin{array}{l}
x=-6+t \cos \alpha, \\
y=3+t \sin \alpha .
\end{array}\right.
$$

Substitute (1) into the parabola equation, eliminate $x$ and $y$ to get $2 t^{2} \sin ^{2} \alpha+(12 \sin \alpha-9 \cos \alpha) t+72=0$.

By the definition of the tangent line and the geometric meaning of $t$, the $t$ in (2) has equal roots, so we have
$$
\begin{array}{l}
(12 \sin \alpha-9 \cos \alpha)^{2}-4 \cdot\left(2 \sin ^{2} \alpha\right) \\
\cdot 72=0 .
\end{array}
$$

Solving (3) yields
$$
\begin{array}{c}
\cos \alpha=-\frac{4}{5}, \\
\sin \alpha=\frac{3}{5}, \\
\cos \alpha=\frac{4}{\sqrt{17}}, \\
\sin \alpha=\frac{1}{\sqrt{17}} .
\end{array}
$$

Substitute (4) and (5) into (1), the parametric equations of the required tangent lines are $\left\{\begin{array}{l}x=-6-\frac{4}{5} t, \\ y=3+\frac{3}{5} t ;\end{array}\right.$
and $\left\{\begin{array}{l}x=-6+\frac{4}{\sqrt{17}} t, \\ y=3+\frac{1}{\sqrt{17}} t .\end{array}\right.$
Convert to general form as $3 x+4 y+6=0$ and $x-4 y+18=0$.