Let $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ be a point on the ellipse.
$\begin{array}{l}
\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{32}=1 \\
\Rightarrow \frac{\mathrm{x}_{1}^{2}}{18}+\frac{\mathrm{y}_{1}^{2}}{32}=1 \ldots \text { (i) }
\end{array}$
The equation of the tangent at $\left(x_{1}, y_{1}\right)$ is $\frac{x x_{1}}{18}+\frac{y y_{1}}{32}=1$. This meets the axes at $\mathrm{A}\left(\frac{18}{\mathrm{x}_{1}}, 0\right)$ and $\mathrm{B}\left(0, \frac{32}{\mathrm{y}_{1}}\right)$. It is given that slope of the tangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is $-\frac{4}{3}$
So, $-\frac{x_{1}}{18} \cdot \frac{32}{y_{1}}=-\frac{4}{3}$
$\Rightarrow \frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{3}{4}$
$\Rightarrow \frac{\mathrm{x}_{1}}{3}=\frac{\mathrm{y}_{1}}{4}=\mathrm{K}$
$\therefore \mathrm{x}_{1}=3 \mathrm{~K}$ and $\mathrm{y}_{1}=4 \mathrm{~K}$
Putting $x_{1}, y_{1}$ in (i), we get
$K^{2}=1$
$\begin{array}{l}
\therefore \text { Area of } \Delta \mathrm{OAB}=\frac{1}{2} \mathrm{OA} . \mathrm{OB} \\
=\frac{1}{2} \cdot \frac{18}{\mathrm{x}_{1}} \cdot \frac{32}{\mathrm{y}_{1}}=\frac{1}{2} \frac{(18)(32)}{(3 \mathrm{~K})(4 \mathrm{~K})}=\frac{24}{\mathrm{~K}^{2}} \\
=24 \text { sq units } \quad\left(\because \mathrm{K}^{2}=1\right)
\end{array}$