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

Given the parametric equations of the curve:

$x = a\sin^3 t$

$y = b\cos^3 t$

At $t = \dfrac{\pi}{2}$, the coordinates of the point on the curve are:

$x = a\sin^3\left(\dfrac{\pi}{2}\right) = a(1)^3 = a$

$y = b\cos^3\left(\dfrac{\pi}{2}\right) = b(0)^3 = 0$

So, the point is $(a, 0)$.

Differentiating $x$ and $y$ with respect to $t$:

$\dfrac{dx}{dt} = 3a\sin^2 t \cos t$

$\dfrac{dy}{dt} = -3b\cos^2 t \sin t$

The slope of the tangent to the curve is:

$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{-3b\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\dfrac{b}{a}\cot t$

At $t = \dfrac{\pi}{2}$, the slope of the tangent is $\dfrac{dy}{dx} = -\dfrac{b}{a}\cot\left(\dfrac{\pi}{2}\right) = 0$.

Since the tangent is horizontal, the normal to the curve is vertical (parallel to the y-axis).

The equation of the vertical normal passing through $(a, 0)$ is:

$x = a \Rightarrow x + 0 \cdot y - a = 0$

Comparing this with the given equation of the normal $x + By + C = 0$, we get:

$B = 0$

$C = -a$

Therefore, the value of $B - C$ is:

$B - C = 0 - (-a) = a$

Answer: $a$