For the curve b{y^2} = {(x + a)^3},the square | vedclass

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(B) Given the curve $b{y^2} = {(x + a)^3}$.Differentiating with respect to $x$,we get $2by \cdot \frac{dy}{dx} = 3{(x + a)^2}$.Thus,$\frac{dy}{dx} = \

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$Subnormal$

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(B) Given the curve $b{y^2} = {(x + a)^3}$.Differentiating with respect to $x$,we get $2by \cdot \frac{dy}{dx} = 3{(x + a)^2}$.Thus,$\frac{dy}{dx} = \frac{3{(x + a)^2}}{2by}$.The formula for the subnormal is $y \cdot \frac{dy}{dx} = y \cdot \frac{3{(x + a)^2}}{2by} = \frac{3{(x + a)^2}}{2b}$.The formula for the subtangent is $\frac{y}{\frac{dy}{dx}} = \frac{y}{\frac{3{(x + a)^2}}{2by}} = \frac{2by^2}{3{(x + a)^2}}$.Substituting $y^2 = \frac{{(x + a)^3}}{b}$ into the subtangent expression:Subtangent $= \frac{2b \cdot \frac{{(x + a)^3}}{b}}{3{(x + a)^2}} = \frac{2(x + a)}{3}$.Therefore,$(Subtangent)^2 = \frac{4}{9}{(x + a)^2}$.Now,consider the ratio $\frac{(Subtangent)^2}{Subnormal} = \frac{\frac{4}{9}{(x + a)^2}}{\frac{3}{2b}{(x + a)^2}} = \frac{8b}{27}$,which is a constant.Hence,$(Subtangent)^2 \propto Subnormal$.

For the curve $b{y^2} = {(x + a)^3}$,the square of the subtangent is proportional to:

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