Correct Answer - Option 1 : -
\(b\over a\) tan t
Concept:
\(\rm d(sinx) \over dt\) = cosx
\(\rm d(cosx) \over dt\)= -sinx
\(\rm sinx \over cosx\)= tanx
Calculation:
Given,
y = b sin3t
\(\rm dy \over dt\)= 3 b sin2t cos t ....(1)
Similarly,
x = a cos3t
\(\rm dx \over dt\) = 3 a cos2t (-sin t) ....(2)
Dividing eq(1) by eq(2) we get,
\(\rm \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)= \(\rm 3 b \sin^2 t \cos t \over -3 a \cos^2 t \sin t\)
⇒\(\rm dy\over dx\) = \(\rm -b \over a \)tan t