A) \[-\frac{4}{9}\]
B) \[-\frac{2}{9}\]
C) \[-\frac{1}{3}\]
D) 1
Correct Answer: A
Solution :
[a] : We have \[{{y}^{2}}={{x}^{3}}\Rightarrow 2y.\frac{dy}{dx}=3{{x}^{2}}\] \[\Rightarrow \frac{dy}{dx}=\frac{3{{x}^{2}}}{2y}\] Given that, slope of tangent at \[({{m}^{2}},{{m}^{3}})\] = slope of normal at \[({{M}^{2}},{{M}^{3}})\] \[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{({{m}^{2}},{{m}^{3}})}}=\frac{-1}{{{\left( \frac{dy}{dx} \right)}_{({{M}^{2}},{{M}^{3}})}}}\] \[\Rightarrow \]\[\frac{3{{m}^{4}}}{2{{m}^{3}}}=-\frac{2{{M}^{3}}}{3{{M}^{4}}}\Rightarrow mM=-\frac{4}{9}\]
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