Question

If the line \(x+By+C = 0\) is the normal to the curve given by \(x = asin^3t\), \(y = bcos^3t\), (where \(a,b\neq 0\)) at a point \(t = \frac{π}{2}\), then \(B-C =\)

• A. \(a\)
• B. \(2a\)
• C. \(-a\)
• D. \(0\)

Hint:

Find the slope of the tangent at t = pi/2 first; it turns out to be zero, so the normal is vertical.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a parametric curve, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). The normal is perpendicular to the tangent, so its slope is \(-\frac{dx}{dy}\) (or it is vertical if the tangent is horizontal).

Step 2: Key Formula or Approach:
\(x = a\sin^3t\) gives \(\frac{dx}{dt} = 3a\sin^2t\cos t\). \(y = b\cos^3t\) gives \(\frac{dy}{dt} = -3b\cos^2t\sin t\).

Step 3: Detailed Explanation:
\[ \frac{dy}{dx} = \frac{-3b\cos^2t\sin t}{3a\sin^2t\cos t} = -\frac{b}{a}\cot t \]
At \(t = \frac\pi2\), \(\cot t = 0\), so the tangent is horizontal and the normal is vertical.
The point is \(x = a\sin^3\frac\pi2 = a\), \(y = b\cos^3\frac\pi2 = 0\), so the normal is \(x = a\), i.e. \(x + 0\cdot y - a = 0\).
Comparing with \(x + By + C = 0\) gives \(B = 0\) and \(C = -a\).
\[ B - C = 0 - (-a) = a \]

Final Answer:
\(B - C = a\), option (A). \[ \boxed{a} \]