question_answer

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:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

A) an ellipse of eccentricity\[\frac{1}{2}\]

B) an ellipse of eccentricity\[\frac{\sqrt{3}}{2}\]

C) a circle of radius\[\frac{1}{2}\]

D)  a circle of radius\[\frac{\sqrt{3}}{2}l\]

Correct Answer: B

Solution :

Let point A (a, 0) is on x-axis and B (0, b) is on y-axis. Let P (h, k) divides AB in the ratio 1 : 2. So, by section formula \[h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3}\] \[k=\frac{2(b)+1(0)}{3}=\frac{2b}{3}\] \[\Rightarrow \]\[a=3h\]and\[b=\frac{3k}{2}\] Now, \[{{a}^{2}}+{{b}^{2}}={{l}^{2}}\]\[\Rightarrow \]\[9{{h}^{2}}+\frac{9{{k}^{2}}}{4}={{l}^{2}}\] \[\Rightarrow \]\[+\frac{{{h}^{2}}}{{{\left( \frac{l}{3} \right)}^{2}}}+\frac{{{k}^{2}}}{{{\left( \frac{2l}{3} \right)}^{2}}}=1\] Now \[e=\sqrt{1-\left( \frac{{{l}^{2}}}{9}\times \frac{9}{4{{l}^{2}}} \right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\] Thus, required locus of P is an ellipse with eccentricity \[\frac{\sqrt{3}}{2}.\]