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 \[{{x}^{2}}+{{y}^{2}}=\frac{{{l}^{2}}}{4}\]You need to login to perform this action.
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