The ratio of the length of the subnormal to t | vedclass

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(A) Given the equation of the curve is $y^2=(2x+1)^3$ ... $(i)$Let $P(x_1, y_1)$ be any point on the curve. So,$y_1^2=(2x_1+1)^3$.Differentiating Eq.

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$27$

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(A) Given the equation of the curve is $y^2=(2x+1)^3$ ... $(i)$Let $P(x_1, y_1)$ be any point on the curve. So,$y_1^2=(2x_1+1)^3$.Differentiating Eq. $(i)$ with respect to $x$,we get:$2y \frac{dy}{dx} = 3(2x+1)^2 	imes 2$$\frac{dy}{dx} = \frac{3(2x+1)^2}{y}$At point $P(x_1, y_1)$,the slope $m = \frac{dy}{dx} = \frac{3(2x_1+1)^2}{y_1}$.Length of subtangent $(ST)$ = $\left| \frac{y_1}{m} ight| = \left| \frac{y_1}{\frac{3(2x_1+1)^2}{y_1}} ight| = \frac{y_1^2}{3(2x_1+1)^2}$.Length of subnormal $(SN)$ = $|y_1 m| = \left| y_1 	imes \frac{3(2x_1+1)^2}{y_1} ight| = 3(2x_1+1)^2$.We need to find the ratio $\frac{SN}{(ST)^2}$:$\frac{SN}{(ST)^2} = \frac{3(2x_1+1)^2}{\left[ \frac{y_1^2}{3(2x_1+1)^2} ight]^2} = \frac{3(2x_1+1)^2 	imes 9(2x_1+1)^4}{y_1^4} = \frac{27(2x_1+1)^6}{y_1^4}$.Since $y_1^2 = (2x_1+1)^3$,then $y_1^4 = (2x_1+1)^6$.Substituting this into the ratio:$\frac{SN}{(ST)^2} = \frac{27(2x_1+1)^6}{(2x_1+1)^6} = 27$.

The ratio of the length of the subnormal to the square of the length of the subtangent at any point $P$ on the curve $y^2=(2x+1)^3$ is

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