The coordinates of the point $\mathrm{P}(a,b)$ lying in the first quadrant on the ellipse $\frac{{\mathrm{x}}^2}{8}+\frac{{\mathrm{y}}^2}{18}=1$,

 so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, 

at P(a,b) then a+b=_________ 

Any point on the ellipse is given by $(\sqrt{8}\cos \mathrm{\theta },\sqrt{18}\sin \mathrm{\theta })$
Now $\frac{2\mathrm{x}}{8}+\frac{2}{18}\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-{\left.\frac{9\mathrm{x}}{4\mathrm{y}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{(\sqrt{8}\cos \mathrm{\theta },\sqrt{18}\sin \mathrm{\theta })}=-\frac{9\sqrt{8}\cos \mathrm{\theta }}{4\sqrt{18}\sin \mathrm{\theta }}=-\frac{\sqrt{9}}{2}\cot \mathrm{\theta }$Hence the equation of the tangent at $(\sqrt{8}\cos \mathrm{\theta },\sqrt{18}\sin \mathrm{\theta })$ is 
$\mathrm{y}-\sqrt{18}\sin \mathrm{\theta }=-\frac{\sqrt{9}}{2}\cot \mathrm{\theta }(\mathrm{x}-\sqrt{8}\cos \mathrm{\theta })$
Therefore, the tangent cuts the coordinates axes at the points $\left(0,\frac{\sqrt{18}}{\sin \mathrm{\theta }}\right)\text{ and }\left(\frac{\sqrt{8}}{\cos \mathrm{\theta }},0\right)$
Thus the area of the triangle formed by this tangent and the coordinate axes is 
$\mathrm{A}=\frac{1}{2}\sqrt{18}\cdot \sqrt{8}\cdot \frac{1}{\cos \mathrm{\theta sin}\mathrm{\theta }}=\frac{6}{\cos \mathrm{\theta sin}\mathrm{\theta }}=12\mathrm{cosec}2\mathrm{\theta }$
But $\mathrm{cosec}2\mathrm{\theta }$ is smallest when $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$ Therefore A is smallest when $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$.
Hence the required point is $\left(\sqrt{8}\frac{1}{\sqrt{2}},\sqrt{18}\frac{1}{\sqrt{2}}\right)=(2,3)$