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Answer: $x y=2$

Given: Let $P(x, y)$ be the point of contact of tangent with curve$y=f(x)$ which intercepts on $\mathrm{x} \text {-axis }=2 x$

To find: We have to show that the equation of curve which pass through $\left ( 1,2 \right )$

Hint: Use equation of tangent at point $P\left ( x,y \right )$ is  $Y-y=\frac{d y}{d x}(X-x)$

Solution: Equation of tangent at $P\left ( x,y \right )$ is  $Y-y=\frac{d y}{d x}(X-x)$

Put Y=0

\begin{aligned} &=>-y=\frac{d y}{d x}(X-x) \\\\ &=>X-x=-y \frac{d x}{d y} \\\\ &=>X=x-y \frac{d x}{d y} \end{aligned}

Co-ordinate of  $B\left(x-y \frac{d x}{d y}, 0\right)$

Given( intercept on x-axis) = 2x

\begin{aligned} &=>x-y \frac{d x}{d y}=2 x \\\\ &=>-y \frac{d x}{d y}=2 x-x \\\\ &=>-y \frac{d x}{d y}=x \end{aligned}

$=>-\frac{d x}{x}=\frac{d y}{y}$        [separating variables]

On integrating on both sides, we get

\begin{aligned} &=>-\int \frac{d x}{x}=\int \frac{d y}{y} \\\\ &=>-\log x=\log y+C \ldots(i) \end{aligned}

It is passing through $\left ( 1,2 \right )$

\begin{aligned} &=>-\log 1=\log 2+C \\\\ &=>C=-\log 2 \end{aligned}

Put $C=-\log 2$ in equation (i) we get

\begin{aligned} &=>-\log x=\log y-\log 2 \\\\ &=>\log x^{-1}=\log \left(\frac{y}{2}\right) \quad\quad\quad\left[-\log x=\log x^{-1}\right] \end{aligned}

\begin{aligned} &=>\log \left(\frac{1}{x}\right)=\log \left(\frac{y}{2}\right) \\\\ &=>\frac{1}{x}=\frac{y}{2} \end{aligned}

Hence, $xy=2$ is required for equation of the curve.

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