At any point for the curve 3y^2 = (x+5)^3,if | vedclass

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(A) Given the curve $3y^2 = (x+5)^3$. Differentiating with respect to $x$,we get $6y \frac{dy}{dx} = 3(x+5)^2$,which implies $\frac{dy}{dx} = \frac{(x

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$8 SN$

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(A) Given the curve $3y^2 = (x+5)^3$. Differentiating with respect to $x$,we get $6y \frac{dy}{dx} = 3(x+5)^2$,which implies $\frac{dy}{dx} = \frac{(x+5)^2}{2y}$. The length of the subtangent $ST = |\frac{y}{dy/dx}| = |\frac{y}{(x+5)^2 / 2y}| = |\frac{2y^2}{(x+5)^2}|$. Substituting $y^2 = \frac{(x+5)^3}{3}$,we get $ST = |\frac{2(x+5)^3}{3(x+5)^2}| = |\frac{2(x+5)}{3}|$. Thus,$(ST)^2 = \frac{4(x+5)^2}{9}$,so $9(ST)^2 = 4(x+5)^2$. The length of the subnormal $SN = |y \frac{dy}{dx}| = |y \cdot \frac{(x+5)^2}{2y}| = |\frac{(x+5)^2}{2}|$. Therefore,$8 SN = 8 \cdot \frac{(x+5)^2}{2} = 4(x+5)^2$. Comparing the two results,we find $9(ST)^2 = 8 SN$.

At any point for the curve $3y^2 = (x+5)^3$,if $ST$ represents the length of the subtangent and $SN$ represents the length of the subnormal,then $9(ST)^2 = $

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