Find the equation of a curve passing through the point $(1,1)$, if the tangent drawn at any point $P(x, y)$ on the curve meets the coordinate axes at $A$ and $B$ such that $P$ is the mid-point of $A B$.
The below figure obtained by the given information
Let the coordinate of the point $P$ is $(x, y)$. It is given that, $P$ is mid-point of $A B$.
So, the coordinates of points $A$ and $B$ are $(2 x, 0)$ and $(0,2 y)$, respectively.
$\therefore \quad$ Slope of $A B=\frac{0-2 y}{2 x-0}=-\frac{y}{x}$
Since, the segment $A B$ is a tangent to the curve at $P$.
$$\begin{array}{ll} \therefore & \frac{d y}{d x}=-\frac{y}{x} \\ \Rightarrow & \frac{d y}{y}=-\frac{d x}{x} \end{array}$$
On integrating both sides, we get
$$\begin{aligned} & \log y=-\log x+\log C \\ & \log y=\log \frac{C}{x}\quad\text{.... (i)} \end{aligned}$$
Since, the given curve passes through $(1,1)$.
$$\begin{array}{ll} \therefore & \log 1=\log \frac{C}{1} \\ \Rightarrow & 0=\log C \\ \Rightarrow & C=1 \end{array}$$
$$\begin{array}{lc} \therefore & \log y=\log \frac{1}{x} \\ \Rightarrow & y=\frac{1}{x} \\ \Rightarrow & x y=1 \end{array}$$
Solve $x \frac{d y}{d x}=y(\log y-\log x+1)$
$$\begin{array}{ll} \text { Given, } & x \frac{d y}{d x}=y(\log y-\log x+1) \\ \Rightarrow & x \frac{d y}{d x}=y \log \left(\frac{y}{x}+1\right) \\ \Rightarrow & \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\quad\text{.... (i)} \end{array}$$
which is a homogeneous equation.
$$\begin{aligned} \text { Put } & \frac{y}{x} =v \text { or } y=v x \\ \therefore & \frac{d y}{d x} =v+x \frac{d v}{d x} \end{aligned}$$
$$\begin{aligned} &\text { On substituting these values in Eq.(i), we get }\\ &\begin{array}{rlrl} \Rightarrow & v+x \frac{d v}{d x} =v(\log v+1) \\ \Rightarrow & x \frac{d v}{d x} =v(\log v+1- \\ \Rightarrow & x \frac{d v}{d x} =v(\log v) \\ \Rightarrow & \frac{d v}{v \log v}=\frac{d x}{x} \end{array} \end{aligned}$$
On integrating both sides, we get
$$\int \frac{d v}{v \log v}=\int \frac{d x}{x}$$
On putting $\log v=u$ in LHS integral, we get
$$\begin{aligned} & \frac{1}{v} \cdot d v=d u \\ & \int \frac{d u}{u}=\int \frac{d x}{x} \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & \log u =\log x+\log C \\ \Rightarrow \quad & \log u =\log C x \\ \Rightarrow \quad & u =C x \\ \Rightarrow \quad & \log v =C x \\ \Rightarrow \quad & \log \left(\frac{y}{x}\right) =C x \end{aligned}$$
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