The smaller area between the ellipse $\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{16}=1$at the line $\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{4}=1$ is

$\mathrm{area} \mathrm{of} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{16}=1 \mathrm{in} \mathrm{first} \mathrm{quadrant} \mathrm{is} \frac{\mathrm{\pi }\left(3\right)\left(4\right)}{4}=3\mathrm{\pi }$

$\mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{line}\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{4}=1 \mathrm{with} \mathrm{coordinate} \mathrm{axes} \mathrm{is} \frac{1}{2}\left(3\right)\left(4\right)=6$

$\mathrm{Required} \mathrm{area}=3\mathrm{\pi }-6$