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(B) Given the curve equation is $y^n = a^{n-1}x$. Differentiating both sides with respect to $x$,we get: $n y^{n-1} \frac{dy}{dx} = a^{n-1}$. Therefor

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$2$

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(B) Given the curve equation is $y^n = a^{n-1}x$. Differentiating both sides with respect to $x$,we get: $n y^{n-1} \frac{dy}{dx} = a^{n-1}$. Therefore,$\frac{dy}{dx} = \frac{a^{n-1}}{n y^{n-1}}$. The length of the subnormal at any point $(x, y)$ is given by the formula: $L = \left| y \frac{dy}{dx} \right|$. Substituting the value of $\frac{dy}{dx}$: $L = \left| y \cdot \frac{a^{n-1}}{n y^{n-1}} \right| = \left| \frac{a^{n-1}}{n y^{n-2}} \right|$. For the length of the subnormal to be constant,it must be independent of $y$. This is possible only if the exponent of $y$ is $0$. Thus,$n - 2 = 0$,which implies $n = 2$.

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

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