The point in the first quadrant of the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{144}=1$ at which the tangent makes equal angels with the axes is (l,m) then l+m=

$$\begin{gathered} \text{ Given ellipse is }\frac{x^2}{25}+\frac{y^2}{144}=1 \\ \text{ Let the point be }\mathrm{P}\equiv \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ which lie on the ellipse } \end{gathered}$$

$\begin{array}{l}\Rightarrow \frac{{\left({\mathrm{x}}_1\right)}^2}{25}+\frac{{\left({\mathrm{y}}_1\right)}^2}{144}=1\\ \Rightarrow \frac{{\mathrm{xx}}_1}{25}+\frac{{\mathrm{yy}}_1}{144}=1\\ \Rightarrow \frac{{\mathrm{yy}}_1}{144}=-\frac{{\mathrm{xx}}_1}{25}+1\\ \Rightarrow \mathrm{y}=-\frac{144{\mathrm{x}}_1}{25{\mathrm{y}}_1}\mathrm{x}+\frac{144}{{\mathrm{y}}_1}\\ \Rightarrow \mathrm{m}=-\frac{144{\mathrm{x}}_1}{25{\mathrm{y}}_1}=\pm 1\\ \frac{{\mathrm{x}}_1}{{\mathrm{y}}_1}=\pm \frac{25}{144}\\ \therefore \frac{{\mathrm{x}}_1^2}{25}+\frac{{\mathrm{y}}_1^2}{144}=1\Rightarrow {\mathrm{x}}_1=\frac{25}{13};{\mathrm{y}}_1=\frac{144}{13}\end{array}$