If the line x + By + C = 0 is the normal to t | vedclass

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(A) Given the curve $x = a \sin^3 t$ and $y = b \cos^3 t$. Find the derivatives: $\frac{dx}{dt} = 3a \sin^2 t \cos t$ and $\frac{dy}{dt} = -3b \cos^2

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$a$

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(A) Given the curve $x = a \sin^3 t$ and $y = b \cos^3 t$. Find the derivatives: $\frac{dx}{dt} = 3a \sin^2 t \cos t$ and $\frac{dy}{dt} = -3b \cos^2 t \sin t$. The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{b}{a} \cot t$. At $t = \frac{\pi}{2}$, $\frac{dy}{dx} = -\frac{b}{a} \cot(\frac{\pi}{2}) = 0$. The slope of the normal is $-\frac{1}{dy/dx}$, which is undefined (vertical line). However, at $t = \frac{\pi}{2}$, the point is $(x, y) = (a \sin^3(\frac{\pi}{2}), b \cos^3(\frac{\pi}{2})) = (a, 0)$. $A$ vertical line passing through $(a, 0)$ is $x = a$, or $x - a = 0$. Comparing $x - a = 0$ with $x + By + C = 0$, we get $B = 0$ and $C = -a$. Therefore, $B - C = 0 - (-a) = a$.

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

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