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A stair-case of length $l$ rests against a vertical wall and a floor of a room. Let $\mathrm{P}$ 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 $\mathrm{P}$ is:
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The correct answer is:
an ellipse of eccentricity $\frac{\sqrt{3}}{2}$
an ellipse of eccentricity $\frac{\sqrt{3}}{2}$
Let point $\mathrm{A}(a, 0)$ is on $x$-axis and $\mathrm{B}$ $(0, b)$ is on $y$-axis.

Let $\mathrm{P}(h, k)$ divides $\mathrm{AB}$ in the ratio $1: 2$. So, by section formula
$$
\begin{aligned}
&h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3} \\
&k=\frac{2(b)+1(0)}{3}=\frac{2 b}{3} \\
&\Rightarrow a=3 h \text { and } b=\frac{3 k}{2}
\end{aligned}
$$
Now, $a^2+b^2=l^2$
$$
\begin{aligned}
&\Rightarrow 9 h^2+\frac{9 k^2}{4}=l^2 \\
&\Rightarrow \frac{h^2}{\left(\frac{l}{3}\right)^2}+\frac{k^2}{\left(\frac{2 l}{3}\right)^2}=1
\end{aligned}
$$
Now $\mathrm{e}=\sqrt{1-\left(\frac{l^2}{9} \times \frac{9}{4 l^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
Thus, required locus of $\mathrm{P}$ is an ellipse with eccentricity $\frac{\sqrt{3}}{2}$.

Let $\mathrm{P}(h, k)$ divides $\mathrm{AB}$ in the ratio $1: 2$. So, by section formula
$$
\begin{aligned}
&h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3} \\
&k=\frac{2(b)+1(0)}{3}=\frac{2 b}{3} \\
&\Rightarrow a=3 h \text { and } b=\frac{3 k}{2}
\end{aligned}
$$
Now, $a^2+b^2=l^2$
$$
\begin{aligned}
&\Rightarrow 9 h^2+\frac{9 k^2}{4}=l^2 \\
&\Rightarrow \frac{h^2}{\left(\frac{l}{3}\right)^2}+\frac{k^2}{\left(\frac{2 l}{3}\right)^2}=1
\end{aligned}
$$
Now $\mathrm{e}=\sqrt{1-\left(\frac{l^2}{9} \times \frac{9}{4 l^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
Thus, required locus of $\mathrm{P}$ is an ellipse with eccentricity $\frac{\sqrt{3}}{2}$.
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