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 staircase begins to slide on the floor, then the locus of P is:

Let us choose a co-ordinate system that makes every relation easy to write. We take the origin $$O(0,0)$$ at the point where the wall meets the floor. The positive $$x$$-axis is along the floor and the positive $$y$$-axis is up the wall.

The foot of the stair-case on the floor is therefore $$F(x,0)$$ and its top on the wall is $$W(0,y)$$. Because the stair-case has fixed length $$l$$, the right-angled triangle $$OFW$$ obeys the Pythagoras theorem

$$x^{2}+y^{2}=l^{2}\;. \quad -(1)$$

Now we locate the special point $$P$$ on the stair-case. The problem states that the segment is divided in the ratio $$1:2$$, and it is nearer to the wall end. That means

$$\dfrac{WP}{PF}= \dfrac{1}{2}\;,$$

so $$WP$$ is one part and $$PF$$ is two parts of the total three parts. Using the section formula for internal division, if a point $$P$$ divides the segment joining $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$ in the ratio $$m:n$$ (with $$m$$ measured from $$A$$), then

$$P\Bigl(\dfrac{nx_{1}+mx_{2}}{m+n},\;\dfrac{ny_{1}+my_{2}}{m+n}\Bigr).$$

Here $$A=W(0,y)$$, $$B=F(x,0)$$, $$m=1$$, $$n=2$$, so

$$P\Bigl(\dfrac{2\cdot0+1\cdot x}{1+2},\;\dfrac{2\cdot y+1\cdot0}{1+2}\Bigr)=\Bigl(\dfrac{x}{3},\;\dfrac{2y}{3}\Bigr).$$

To find the locus we rename the co-ordinates of $$P$$ as $$P(X,Y)$$, that is

$$X=\dfrac{x}{3},\qquad Y=\dfrac{2y}{3}.$$

We now express $$x$$ and $$y$$ through $$X$$ and $$Y$$:

$$x=3X,\qquad y=\dfrac{3}{2}\,Y.$$

Substituting these values in equation (1) we have

$$\bigl(3X\bigr)^{2}+\Bigl(\dfrac{3}{2}Y\Bigr)^{2}=l^{2}.$$

That simplifies step by step as follows:

$$9X^{2}+\dfrac{9}{4}Y^{2}=l^{2}$$

$$\Longrightarrow\; \dfrac{9X^{2}}{l^{2}}+\dfrac{9Y^{2}}{4l^{2}}=1$$

$$\Longrightarrow\; \dfrac{X^{2}}{l^{2}/9}+\dfrac{Y^{2}}{4l^{2}/9}=1.$$

This is clearly the canonical form of an ellipse

$$\dfrac{X^{2}}{a^{2}}+\dfrac{Y^{2}}{b^{2}}=1$$

with

$$a^{2}=\dfrac{l^{2}}{9},\qquad b^{2}=\dfrac{4l^{2}}{9},\qquad b^{2}>a^{2}.$$

For an ellipse with $$b>a$$, the eccentricity $$e$$ is defined by

$$e=\sqrt{1-\dfrac{a^{2}}{b^{2}}}.$$

Hence

$$e=\sqrt{1-\dfrac{l^{2}/9}{4l^{2}/9}}=\sqrt{1-\dfrac{1}{4}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}.$$

Thus, as the stair-case slides, the point $$P$$ traces an ellipse of eccentricity $$\dfrac{\sqrt{3}}{2}$$.

Hence, the correct answer is Option B.