The curve for which the intercept cut off by | vedclass

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Question

(C) Let the point of tangency be $(x, y)$. The equation of the tangent line is $Y - y = \frac{dy}{dx}(X - x)$.To find the $y$-intercept,set $X = 0$:$Y

Vedclass · Accepted Answer

$\frac{c_1}{x} + \frac{c_2}{y} = 1$

Vedclass · Answer

(C) Let the point of tangency be $(x, y)$. The equation of the tangent line is $Y - y = \frac{dy}{dx}(X - x)$.To find the $y$-intercept,set $X = 0$:$Y - y = \frac{dy}{dx}(0 - x) \implies Y = y - x \frac{dy}{dx}$.According to the problem,the $y$-intercept $Y$ is proportional to the square of the ordinate $y$,so $Y = ky^2$ for some constant $k$.Equating the two expressions for $Y$:$y - x \frac{dy}{dx} = ky^2$.Rearranging the terms:$x \frac{dy}{dx} - y = -ky^2$.Divide by $xy^2$:$\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{xy} = -\frac{k}{x}$.Let $v = \frac{1}{y}$,then $\frac{dv}{dx} = -\frac{1}{y^2} \frac{dy}{dx}$.Substituting this into the equation:$-\frac{dv}{dx} - \frac{v}{x} = -\frac{k}{x} \implies \frac{dv}{dx} + \frac{v}{x} = \frac{k}{x}$.This is a linear differential equation. The integrating factor is $e^{\int \frac{1}{x} dx} = x$.Multiplying by $x$:$x \frac{dv}{dx} + v = k \implies \frac{d}{dx}(xv) = k$.Integrating both sides:$xv = kx + C \implies x(\frac{1}{y}) = kx + C$.Dividing by $x$:$\frac{1}{y} = k + \frac{C}{x} \implies \frac{C}{x} + \frac{1}{y} = k$.Rearranging into the form $\frac{c_1}{x} + \frac{c_2}{y} = 1$ by dividing by $k$ (where $c_1 = C/k$ and $c_2 = 1/k$).

The curve for which the intercept cut off by any tangent on the $y$-axis is proportional to the square of the ordinate of the point of tangency is (where $c_1$ and $c_2$ are arbitrary constants):

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