If a curve passes through the origin,such tha | vedclass

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(A) The length of the subnormal is given by $y \frac{dy}{dx}$.Given that the length of the subnormal is one more than the square of the ordinate $(y^2

Vedclass · Accepted Answer

$f(x) = \sqrt{e^{2x} - 1}$

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(A) The length of the subnormal is given by $y \frac{dy}{dx}$.Given that the length of the subnormal is one more than the square of the ordinate $(y^2)$,we have the differential equation:$y \frac{dy}{dx} = y^2 + 1$Rearranging the terms,we get:$\frac{y}{y^2 + 1} dy = dx$Integrating both sides:$\int \frac{y}{y^2 + 1} dy = \int dx$$\frac{1}{2} \ln(y^2 + 1) = x + C$Since the curve passes through the origin $(0, 0)$,we substitute $x=0$ and $y=0$:$\frac{1}{2} \ln(0^2 + 1) = 0 + C \implies C = 0$Thus,$\frac{1}{2} \ln(y^2 + 1) = x$$\ln(y^2 + 1) = 2x$$y^2 + 1 = e^{2x}$$y^2 = e^{2x} - 1$$y = \sqrt{e^{2x} - 1}$Therefore,the correct option is $A$.

If a curve passes through the origin,such that the length of the subnormal is equal to one more than the square of the ordinate,then:

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