If the length of the sub-tangent at any point | vedclass

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(C) We know that the length of the sub-tangent is given by $\frac{y}{dy/dx}$.Given, $\frac{y}{dy/dx} = x + 7y^2$.Rearranging the terms, we get $\frac{

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$7y^2 + cy - x$

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(C) We know that the length of the sub-tangent is given by $\frac{y}{dy/dx}$.Given, $\frac{y}{dy/dx} = x + 7y^2$.Rearranging the terms, we get $\frac{dy}{dx} = \frac{y}{x + 7y^2}$.Taking the reciprocal, we have $\frac{dx}{dy} = \frac{x + 7y^2}{y} = \frac{x}{y} + 7y$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = -\frac{1}{y}$ and $Q(y) = 7y$.The integrating factor $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$.The solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.$x \cdot \frac{1}{y} = \int 7y \cdot \frac{1}{y} dy + C$.$\frac{x}{y} = \int 7 dy + C = 7y + C$.$x = 7y^2 + Cy$.Thus, $7y^2 + Cy - x = 0$, which corresponds to $f(x, y) = 7y^2 + cy - x$.

If the length of the sub-tangent at any point $P(x, y)$ on a curve $f(x, y) = 0$ is $x + 7y^2$, then $f(x, y) =$

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