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(C) Let the point of contact be $(x, y)$. The equation of the tangent at $(x, y)$ is given by $(Y - y) = \frac{dy}{dx}(X - x)$.To find the $y$-interce

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$x^2y=C$,where $C$ is a constant

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(C) Let the point of contact be $(x, y)$. The equation of the tangent at $(x, y)$ is given by $(Y - y) = \frac{dy}{dx}(X - x)$.To find the $y$-intercept,we set $X = 0$ in the tangent equation:$Y - y = \frac{dy}{dx}(0 - x) \Rightarrow Y = y - x \frac{dy}{dx}$.The length of the $y$-intercept is $|Y| = |y - x \frac{dy}{dx}|$. According to the problem,the $y$-intercept is $3$ times the ordinate $y$:$y - x \frac{dy}{dx} = 3y$.Rearranging the terms,we get $-x \frac{dy}{dx} = 2y$.Separating the variables,we have $\frac{dy}{y} = -2 \frac{dx}{x}$.Integrating both sides,we get $\int \frac{dy}{y} = -2 \int \frac{dx}{x} + \ln C$.$\ln y = -2 \ln x + \ln C \Rightarrow \ln y = \ln(x^{-2}) + \ln C$.$\ln y = \ln(C x^{-2}) \Rightarrow y = \frac{C}{x^2}$.Thus,$x^2y = C$,where $C$ is a constant.

$A$ family of curves is such that the length intercepted on the $y$-axis between the origin and the tangent at any point $(x, y)$ is three times the ordinate of the point of contact. The family of curves is:

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