A) \[\frac{t(2-{{t}^{3}})}{(1-2{{t}^{3}})}\]
B) \[\frac{t(2+{{t}^{3}})}{(1-2{{t}^{3}})}\]
C) \[\frac{t(2-{{t}^{3}})}{(1+2{{t}^{3}})}\]
D) \[\frac{t(2+{{t}^{3}})}{(1+2{{t}^{3}})}\]
Correct Answer: A
Solution :
Given, \[y=\frac{3a{{t}^{2}}}{1+{{t}^{3}}},x=\frac{3at}{1+{{t}^{3}}}\] On differentiating w.r.t.t respectively, we get \[\frac{dy}{dx}=\frac{(1+{{t}^{3}})\,(6at)-3a{{t}^{2}}(3{{t}^{2}})}{{{(1+{{t}^{3}})}^{2}}}\] \[=\frac{6at-3a{{t}^{4}}}{{{(1+{{t}^{3}})}^{2}}}\] and \[\frac{dx}{dt}=\frac{(1-{{t}^{3}})\,(3a)-3at\,(3{{t}^{2}})}{{{(1+{{t}^{3}})}^{2}}}\] \[=\frac{3a-6a{{t}^{2}}}{{{(1+{{t}^{3}})}^{2}}}\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3at\,(2-{{t}^{3}})}{3a(1-2{{t}^{3}})}\] \[=\frac{t(2-{{t}^{3}})}{(1-2{{t}^{3}})}\]
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