Question

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 :

• A. A circle of radius $\frac{l}{2}$
• B. An ellipse of eccentricity $\frac{1}{2}$
• C. A circle of radius $\frac{\sqrt{3}}{2}l$
• D. An ellipse of eccentricity $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D An ellipse of eccentricity $\frac{\sqrt{3}}{2}$

Let b be the height and a be the length intercepted by the staircase.
By section formula, we can write the coordinates of $P$ as :
$(\frac{b}{3},\frac{2a}{3})$
Now, the length of the staircase is constant.
Hence, $a^2+b^2=l^2$
Let $P$ be $(x,y)$.
Hence,
$(3x{)}^2+{\left(\frac{3y}{2}\right)}^2=l^2$
$\Rightarrow \frac{x^2}{\frac{l^2}{9}}+\frac{y^2}{\frac{4l^2}{9}}=1$
This represents the equation of an ellipse ,
$1-e^2=\frac{1}{4}$
Hence, $e=\frac{\sqrt{3}}{2}$