A curve passes through the point (3,2) for wh | vedclass

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(C) Let the point of contact be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$, where $m = \frac{dy}{dx}$.Since t

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$xy=6$

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(C) Let the point of contact be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$, where $m = \frac{dy}{dx}$.Since the segment of the tangent between the coordinate axes is bisected at $(x_1, y_1)$, the points where the tangent meets the axes are $(2x_1, 0)$ and $(0, 2y_1)$.The slope of the tangent is $m = \frac{2y_1 - 0}{0 - 2x_1} = -\frac{y_1}{x_1}$.Thus, $\frac{dy}{dx} = -\frac{y}{x}$.Separating variables, we get $\frac{dy}{y} = -\frac{dx}{x}$.Integrating both sides, $\ln y = -\ln x + C_0$, which implies $\ln(xy) = C_0$, or $xy = C$.Since the curve passes through $(3, 2)$, we have $3 	imes 2 = C$, so $C = 6$.Therefore, the equation of the curve is $xy = 6$.

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

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