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(B) Given the curve $y = a^{1-n}x^n$. First,we find the derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = n a^{1-n} x^{n-1}$. The length of the subnormal a

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$1/2$

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(B) Given the curve $y = a^{1-n}x^n$. First,we find the derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = n a^{1-n} x^{n-1}$. The length of the subnormal at any point $(x, y)$ is given by the formula $L = |y \cdot \frac{dy}{dx}|$. Substituting the values: $L = |(a^{1-n}x^n) \cdot (n a^{1-n} x^{n-1})| = |n a^{2(1-n)} x^{2n-1}|$. For the length of the subnormal to be constant,the exponent of $x$ must be zero. Therefore,$2n - 1 = 0$,which implies $n = 1/2$.

If the length of the subnormal at any point on the curve $y = a^{1-n}x^n$ is constant,then $n = \dots$

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