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: Use the point and direction:
At $t = \frac\pi2$ the point is $(a, 0)$. The tangent direction is $(dx, dy) \propto (3a\sin^2t\cos t,\ -3b\cos^2t\sin t) = (0, 0)$ at that parameter value, so use the limit $\frac{dy}{dx} = -\frac ba\cot t \to 0$.

Step 2: Write the normal:
A horizontal tangent means the normal is the vertical line through $(a,0)$: $x = a$.
In the form $x + By + C = 0$: $B = 0$, $C = -a$, so $B - C = a$.

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