A) \[\frac{1}{3}\]
B) \[\frac{1}{3},3\]
C) \[\frac{2}{3},3\]
D) \[3\]
Correct Answer: D
Solution :
Let \[\frac{AN}{B\lambda }=\lambda \] then by section formula, the coordinates of N are \[\left( \frac{a}{1+\lambda },\frac{\lambda a}{1+\lambda } \right)\] where (a, 0) and (0, a) are the coordinates of A and B respectively. Since, slope of \[AB=\frac{0-a}{a-0}=-1\] \[\therefore \] Slope of \[MN=1\] \[(\because \,MN\bot AB)\] Now, equation of MN \[y-\frac{\lambda a}{1+\lambda }=x-\frac{a}{1+\lambda }\] \[\Rightarrow \] \[x-y=\frac{1-\lambda }{1+\lambda }a\] By putting \[x=0\]in this equation, we get the coordinates of M which are \[\left[ 0,\left( \frac{\lambda -1}{\lambda +1} \right)a \right].\] Now, it is given that area of\[\Delta AMN=\frac{3}{8}.\]area of \[\Delta OAB\] \[\Rightarrow \] \[\frac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \\ a & 0 & \frac{a}{1+\lambda } \\ 0 & \left( \frac{\lambda -1}{\lambda +1} \right) & \frac{\lambda a}{1+\lambda } \\ \end{matrix} \right|=\frac{3}{8}.\frac{1}{2}{{a}^{2}}\] \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] \[\Rightarrow \] \[\frac{1}{2}\left| \begin{matrix} 1 & 0 & 0 \\ a & -a & \frac{-a\lambda }{1+\lambda } \\ 0 & \left( \frac{\lambda -1}{\lambda +1} \right) & \frac{\lambda a}{1+\lambda } \\ \end{matrix} \right|=\frac{3}{16}{{a}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}\left| -a\left( \frac{\lambda a}{1+\lambda } \right)+\frac{a\lambda }{1+\lambda }\left( \frac{\lambda -1}{\lambda +1} \right)a \right|=\frac{3}{16}{{a}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}\lambda {{a}^{2}}\left| \frac{\lambda -1}{{{(\lambda +1)}^{2}}}-\frac{1}{\lambda +1} \right|=\frac{3}{16}{{a}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}\lambda {{a}^{2}}\left| \frac{-2}{{{(\lambda +1)}^{2}}} \right|=\frac{3}{16}{{a}^{2}}\] \[\Rightarrow \] \[\frac{\lambda {{a}^{2}}}{{{(\lambda +1)}^{2}}}=\frac{3}{16}{{a}^{2}}\] \[\Rightarrow \] \[16\lambda =3{{\lambda }^{2}}+3+6\lambda \] \[\Rightarrow \] \[3{{\lambda }^{2}}-10\lambda +3=0\] \[\Rightarrow \] \[(\lambda -3)(3\lambda -1)=0\] \[\Rightarrow \] \[\lambda =3\]or \[\lambda =\frac{1}{3}\] For \[\lambda =\frac{1}{3}.\,M\]lies outside the segment OB and hence the required value of \[\lambda \]is 3. ie, \[\frac{AN}{BN}=3\]You need to login to perform this action.
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