A) \[x\sec t-y\cos ect=\alpha \]
B) \[x\sec t+y\cos ect=\alpha \]
C) \[x\cos ect+ysect=\alpha \]
D) None of the above
Correct Answer: B
Solution :
\[\frac{dy}{dx}=3a{{\sin }^{2}}t.\cos t,\] \[\frac{dx}{dt}=-3a{{\cos }^{2}}t.\sin t\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{\sin t}{\cos t}\] \[\therefore \]Equation of tangent is \[x\sin t+u\cos t=a\sin t.\cos t({{\cos }^{2}}t+{{\sin }^{2}}t)\] \[\Rightarrow \] \[x\text{ }sin\text{ }t+y\text{ }cost=a\text{ }sin\text{ }t.cost\] \[\Rightarrow \] \[x\text{ }sec\text{ }t+\text{ }cosec\text{ }t=0\]
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