A) \[\left( 3a,\frac{27}{4}a \right)\]
B) \[(2a,8a)\]
C) \[(a,0)\]
D) None of these
Correct Answer: A
Solution :
The ordinate of any point on the curve is given by \[y=\frac{{{x}^{3}}}{{{(x-a)}^{2}}}\] \[\frac{dy}{dx}=\frac{3{{x}^{2}}}{{{(x-a)}^{2}}}-\frac{2{{x}^{3}}}{{{(x-a)}^{3}}}=\frac{{{x}^{2}}(x-3a)}{{{(x-a)}^{3}}},\] Now, \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[x=0\]or \[x=3a\] And \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{(x-a)}^{3}}(3{{x}^{2}}-6ax)-3{{x}^{2}}(x-3a){{(x-a)}^{2}}}{{{(x-a)}^{6}}}\] \[{{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=0}}=0\] And \[{{\left. \frac{{{d}^{2}}y}{d{{x}^{2}}} \right|}_{x=3a}}=\frac{8{{a}^{3}}(9{{a}^{2}})-27{{a}^{2}}\times 0}{{{(2a)}^{6}}}>0\] Hence, \[y\]is minimum at \[x=3a\]and its value is \[y=\frac{{{(3a)}^{3}}}{{{(2a)}^{2}}}=\frac{27{{a}^{2}}}{4{{a}^{2}}}=\frac{27}{4}a.\]
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