A stair – case of length $l$ rests against a vertical wall and a floor of a room. Let $P$be a point on the stair-case, nearer to its end on the wall, that divides its length in the ratio $1 : 2$. If the stair-case begins to slide on the floor, then the locus of $P$is

Let point $A (a,0)$ is on $x-axis$ and $B(0,b)$ is on $y-axis$.

$$\begin{gathered} \mathrm{Let} P(h,k), \mathrm{divides} AB \mathrm{in} \mathrm{the} \mathrm{ratio} 2 : 1. So, \mathrm{by} \mathrm{section} \mathrm{formula}, \\ If a point Q(x , y) divides a line segment joining C(x_1 , y_{ }) , D(x_2 ,y_2) in the ratio m : n internally, \\ Then, Q(x ,y) = (\frac{m{x}_2+n{x}_1}{m+n}, \frac{m{y}_2+n{y}_1}{m+n}). \\ So, u\sin g the above formula, \\ \mathrm{h}=\frac{2\left(0\right)+1\left(\mathrm{a}\right)}{1+2}=\frac{\mathrm{a}}{3} \\ \mathrm{k}=\frac{2\left(\mathrm{b}\right)+1\left(0\right)}{3}=\frac{2\mathrm{b}}{3} \\ \Rightarrow \mathrm{a}=3\mathrm{h} \mathrm{and} \mathrm{b}=\frac{3\mathrm{k}}{2} \\ \mathrm{Now}, {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{l}}^2 \\ \Rightarrow 9{\mathrm{h}}^2+\frac{9{\mathrm{k}}^2}{4}={\mathrm{l}}^2 \\ \Rightarrow \frac{{\mathrm{h}}^2}{{\left(\frac{\mathrm{l}}{3}\right)}^2}+\frac{{\mathrm{k}}^2}{{\left({\displaystyle \frac{2\mathrm{l}}{3}}\right)}^2}=1 \\ \mathrm{Now} \mathrm{e}=\sqrt{1-\left(\frac{{\mathrm{l}}^2}{9}\times \frac{9}{4{\mathrm{l}}^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} (Eccentricity of ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is \sqrt{\frac{b^2-a^2}{b^2}} if b > a) \end{gathered}$$

Thus, required locus of $P$is an ellipse with eccentricity$\frac{\sqrt{3}}{2}$

Hence, the correct option is 2.