The subnormal at any point on the curve xy^n | vedclass

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(C) The equation of the curve is $xy^n = a^{n+1}$.Taking the derivative with respect to $x$:$y^n + x \cdot n y^{n-1} \frac{dy}{dx} = 0$$x n y^{n-1} \f

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$-2$

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(C) The equation of the curve is $xy^n = a^{n+1}$.Taking the derivative with respect to $x$:$y^n + x \cdot n y^{n-1} \frac{dy}{dx} = 0$$x n y^{n-1} \frac{dy}{dx} = -y^n$$\frac{dy}{dx} = -\frac{y^n}{n x y^{n-1}} = -\frac{y}{nx}$.The length of the subnormal is given by $|y \frac{dy}{dx}|$.Substituting $\frac{dy}{dx}$:$|y \cdot (-\frac{y}{nx})| = |-\frac{y^2}{nx}|$.From the original equation,$y^n = \frac{a^{n+1}}{x}$,so $y^2 = (\frac{a^{n+1}}{x})^{2/n} = \frac{a^{2(n+1)/n}}{x^{2/n}}$.Substituting this into the subnormal expression:$|\frac{y^2}{nx}| = |\frac{a^{2(n+1)/n}}{n x \cdot x^{2/n}}| = |\frac{a^{2(n+1)/n}}{n x^{(n+2)/n}}|$.For the subnormal to be constant,the power of $x$ must be $0$,so $\frac{n+2}{n} = 0$,which implies $n = -2$.Thus,for $n = -2$,the subnormal is constant.

The subnormal at any point on the curve $xy^n = a^{n + 1}$ is constant for $n=$ ............

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