question_answer

A rod of length \[l\]slides with its ends of two perpendicular lines. Then, the locus of its mid-point is

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

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

C)  \[{{x}^{2}}-{{y}^{2}}=\frac{{{l}^{2}}}{4}\]

D)  None of these

Correct Answer: A

Solution :

Let both of the ends of the rod are on \[x\text{-}\]axis and \[y\text{-}\]axis. Let AB be rod of length \[l\]and coordinates of A and B be (a, 0) and (0, b) respectively. Let P (h, k) be the midpoint of the rod AB. Then, \[\left. \begin{align}   & h=\frac{0+a}{2}=\frac{a}{2} \\  & k=\frac{b+0}{2}=\frac{b}{2} \\ \end{align} \right\}\] ?(i) Now, in \[\Delta OAB,\] \[O{{A}^{2}}+O{{B}^{2}}=A{{B}^{2}}\] \[{{a}^{2}}+{{b}^{2}}={{l}^{2}}\] \[\Rightarrow \] \[{{(2h)}^{2}}+{{(2k)}^{2}}={{l}^{2}}\] [using Eq. (i)] \[\Rightarrow \] \[{{h}^{2}}+{{k}^{2}}=\frac{{{l}^{2}}}{4}\] \[\therefore \] The equation of locus is \[{{x}^{2}}+{{y}^{2}}=\frac{{{l}^{2}}}{4}\]