Step 1: Checking the point directly:
At \(t=\dfrac{\pi}{2}\): \(x=a\sin^3\dfrac{\pi}{2}=a\), \(y=b\cos^3\dfrac{\pi}{2}=0\), so the curve is at \((a,0)\).
Step 2: Local behaviour of y near t=π/2:
Near \(t=\pi/2\), \(\cos t\approx -(t-\pi/2)\) is small and to the first power \(y=b\cos^3t\) vanishes to third order, while \(x=a\sin^3t\) changes only to second order in \((t-\pi/2)\) since \(\sin t\approx1-\tfrac12(t-\pi/2)^2\) — so \(y\) falls to zero much faster than \(x\) departs from \(a\), meaning the slope of the curve there is \(0\).
Step 3: Confirming via the formula:
This matches \(\dfrac{dy}{dx}=-\dfrac{b}{a}\cot t\), which is \(0\) exactly at \(t=\pi/2\) since \(\cot(\pi/2)=0\).
Final Answer:
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