A) \[{{x}^{3}}\]
B) \[{{x}^{2}}\]
C) \[{{y}^{3}}\]
D) \[\infty \]
Correct Answer: C
Solution :
We have \[xy={{c}^{2}}\] \[\therefore \] \[y=\frac{{{c}^{2}}}{x}\] ... (i) \[\therefore \] \[\frac{dy}{dx}={{c}^{2}}\left( -\frac{1}{{{x}^{2}}} \right)\] \[\therefore \] Sub normal at any point \[=y\,.\,\frac{dy}{dx}\] \[=y\times \left( -\frac{{{c}^{2}}}{{{x}^{2}}} \right)\] \[=y\times \frac{-{{c}^{2}}}{{{\left( \frac{{{c}^{2}}}{y} \right)}^{2}}}\] \[\left[ x=\frac{{{c}^{2}}}{y} \right]\] \[=y\times \frac{-{{c}^{2}}{{y}^{2}}}{{{c}^{4}}}\] \[=-\frac{{{y}^{3}}}{{{c}^{2}}}\] \[\therefore \] Subnormal \[\propto {{y}^{3}}\]
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