For the curve {y^n} = {a^{n - 1}}x,the subnor | vedclass

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

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$2$

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(A) Given the curve equation: ${y^n} = {a^{n - 1}}x$. Differentiating both sides with respect to $x$,we get: $n{y^{n - 1}}\frac{{dy}}{{dx}} = {a^{n - 1}}$. Thus,the slope of the tangent is: $\frac{{dy}}{{dx}} = \frac{{{a^{n - 1}}}}{{n{y^{n - 1}}}}$. The length of the subnormal is given by the formula: $	ext{Subnormal} = y \left| \frac{{dy}}{{dx}} ight|$. Substituting the value of $\frac{{dy}}{{dx}}$: $	ext{Subnormal} = y \cdot \frac{{{a^{n - 1}}}}{{n{y^{n - 1}}}} = \frac{{{a^{n - 1}}}}{n} \cdot y^{1 - (n - 1)} = \frac{{{a^{n - 1}}}}{n} \cdot y^{2 - n}$. For the subnormal to be constant,it must be independent of $y$. This implies the exponent of $y$ must be $0$,so $2 - n = 0$,which gives $n = 2$.

For the curve ${y^n} = {a^{n - 1}}x$,the subnormal at any point is constant. The value of $n$ must be

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