A stair-case of length $l$ rests against a vertical wall and a floor of a room. Let $\mathrm{P}$ be a point…

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:
  1. an ellipse of eccentricity $\frac{1}{2}$
  2. an ellipse of eccentricity $\frac{\sqrt{3}}{2}$
  3. a circle of radius $\frac{1}{2}$
  4. a circle of radius $\frac{\sqrt{3}}{2} l$

Solution

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 $\frac{\sqrt{3}}{2}$

Asked in: JEE Main 2014 (11 Apr Online)

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