The ends of a rod of length move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio $1:2$, is 

Let the equation of the rod be $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$
Where ${\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{l}}^2$........(1)
The point $\mathrm{A}(\mathrm{a},0)$ is on x-axis and $\mathrm{B}(0,\mathrm{b})$ is on y-axis. Let  divide AB in the ratio $\mathrm{C}(\mathrm{h},\mathrm{k})$. So, by section formula,  
$\mathrm{h}=\frac{2\times 0+1\times \mathrm{a}}{2+1}=\frac{\mathrm{a}}{3},$ $\therefore \mathrm{a}=3\mathrm{h}$ and $\mathrm{k}=\frac{2\times \mathrm{b}\times 1\times 0}{2+1}=\frac{2\mathrm{b}}{3},\therefore \mathrm{b}=\frac{3\mathrm{k}}{2}$
on putting for a and b in equation $\left(1\right),$ we get 
$9{\mathrm{h}}^2+9\frac{{\mathrm{k}}^2}{4}={\mathrm{l}}^2$     So, the required locus is $\frac{{\mathrm{x}}^2}{1}+\frac{{\mathrm{y}}^2}{4}=\frac{{\mathrm{l}}^2}{9}.$