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A curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the co-ordinate axes is bisected at the point of contact. The equation of the curve is
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$x y=6$
$\frac{x}{x_1}+\frac{y}{y_1}=2$
$\because\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=-\frac{y_1}{x_1}$
$\therefore \frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}=-\frac{d x}{x} \Rightarrow \ln y=-\ln x+k \Rightarrow \ln x y=k \Rightarrow x y=e^k=C \Rightarrow x y=C$ passes through $(3,2)$
$\therefore C=3 \times 2=6 \therefore x y=6$
$\because\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=-\frac{y_1}{x_1}$
$\therefore \frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}=-\frac{d x}{x} \Rightarrow \ln y=-\ln x+k \Rightarrow \ln x y=k \Rightarrow x y=e^k=C \Rightarrow x y=C$ passes through $(3,2)$
$\therefore C=3 \times 2=6 \therefore x y=6$
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