question_answer

The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is [IIT 1987; RPET 1997]

A) \[36{{x}^{2}}+9{{y}^{2}}=4{{l}^{2}}\]

B) \[36{{x}^{2}}+9{{y}^{2}}={{l}^{2}}\]

C) \[9{{x}^{2}}+36{{y}^{2}}=4{{l}^{2}}\]

D) None of these

Correct Answer: C

Solution :

According to the figure \[AP\,\,:\,\,PB=1\,\,:\,\,2,\] then \[h=\frac{1\times 0+2\times a}{1+2}=\frac{2a}{3}\] or \[a=\frac{3h}{2},\] similarly \[b=3k\] Now we have \[O{{A}^{2}}+O{{B}^{2}}=A{{B}^{2}}\] \[\Rightarrow \,\,{{\left( \frac{3h}{2} \right)}^{2}}+{{(3k)}^{2}}={{l}^{2}}\] Hence locus of \[P(h,\,\,k)\] is given by \[9{{x}^{2}}+36{{y}^{2}}=4{{l}^{2}}\].