A) \[-\frac{b}{a}\tan t\]
B) \[-\frac{b}{a}\cot t\]
C) \[-\frac{a}{b}\cot t\]
D) \[-\frac{a}{b}\tan t\]
Correct Answer: B
Solution :
\[x=a\cos t,y=b\sin t\] \[\therefore \] \[\frac{dx}{dt}=-a\sin t,\] and \[\frac{dy}{dt}=b\cos t\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy/dx}{dx/dt}\] \[=\frac{b\cos t}{-a\sin t}=-\frac{b}{a}\cot t\]
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