A) \[ax\]
B) \[{{a}^{2}}{{x}^{2}}\]
C) \[\frac{x}{a}\]
D) \[\frac{x}{2a}\]
E) x\[2a\]
Correct Answer: C
Solution :
\[\because \]\[x=\frac{2at}{1+{{t}^{3}}}\]and\[y=\frac{2a{{t}^{2}}}{{{(1+{{t}^{3}})}^{2}}}\] \[\Rightarrow \] \[y=2a{{\left( \frac{t}{1+{{t}^{3}}} \right)}^{2}}=2a.\frac{{{x}^{2}}}{{{(2a)}^{2}}}\] \[\Rightarrow \] \[y=\frac{{{x}^{2}}}{2a}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{2x}{2a}=\frac{x}{a}\]
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