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

let $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1,$ be the line which cuty x-axis at A=(a, 0)and y-axis at B=(0, b)

Given that AB=l

$$\begin{gathered} \Rightarrow \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\mathrm{l} \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{l}}^2 - \left(1\right) \end{gathered}$$

Let p(x, y) be locus point which divides $\overline{\mathrm{AB}}$ in the ratio 1 : 2

$$\begin{gathered} \mathrm{P}=\left[\frac{1\left(0\right)+2\left(\mathrm{a}\right)}{3}, \frac{1\left(\mathrm{b}\right)+2\left(0\right)}{3}\right] \\ (\mathrm{x} ,\mathrm{y})=\left(\frac{2\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right) \\ \Rightarrow \mathrm{x}=\frac{2\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3} \\ \Rightarrow \mathrm{a}=\frac{3\mathrm{x}}{2}, \mathrm{b}=3\mathrm{y} \\ \mathrm{from} \left(1\right) \Rightarrow \frac{9{\mathrm{x}}^2}{4}+9{\mathrm{y}}^2={\mathrm{l}}^2 \\ \Rightarrow 9{\mathrm{x}}^2+36{\mathrm{y}}^2=4{\mathrm{l}}^2 \end{gathered}$$