A straight line meets the x and y axes at the points A, B respectively. If AB = 6 units, then the locus of the point P which divides the line segment AB such that AP:PB=2:1 is

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

Given that AB=6

$$\begin{gathered} \Rightarrow \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=6 \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=36 - \left(1\right) \end{gathered}$$

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

$$\begin{gathered} \mathrm{P}=\left[\frac{2\left(0\right)+1\left(\mathrm{a}\right)}{3}, \frac{2\left(\mathrm{b}\right)+1\left(0\right)}{3}\right] \\ \Rightarrow \left[\mathrm{x}, \mathrm{y}\right]=\left[\frac{\mathrm{a}}{3}, \frac{2\mathrm{b}}{3}\right] \\ \Rightarrow \mathrm{a}=3\mathrm{x}, \mathrm{b}=\frac{3\mathrm{y}}{2} \\ \mathrm{from} \left(1\right) \Rightarrow 9{\mathrm{x}}^2+\frac{9{\mathrm{y}}^2}{4}=36 \\ \Rightarrow 9\left[\frac{4{\mathrm{x}}^2+{\mathrm{y}}^2}{4}\right]=36 \\ \Rightarrow 4{\mathrm{x}}^2+{\mathrm{y}}^2=16 \end{gathered}$$