**Toolbox:**

- If a linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x)$ then it can be solved by seperating the variable and then integrating.
- $\large \int \frac{dx}{x}$$=\log x$

Let $P(x,y)$ be any point on the curve $y=f(x)$

Such that the tangent at P cuts the coordinate axes at A and B. The coordinate of A and B are $(x-y \large\frac{dy}{dx}$$,0)$ and $(0, y-x \large\frac{dy}{dx})$ respectively

It is given that $P(x,y)$ is the mid-point of $AB$

Therefore $\large\frac{x-y \Large\frac{dy}{dx}\large+0}{2}$$=x=>x-y \large\frac{dy}{dx}$$=2x$

or $x=-y \large\frac{dy}{dx}$

and $\large\frac{0+y -x\Large\frac{dy}{dx}}{2}$$=y=>y-x \large\frac{dy}{dx}$$=2y$

or $y=-x \large\frac{dy}{dx}$

Consider or $y=-x \large\frac{dy}{dx}$

Now seperating the variables,

$\large\frac{dy}{y}=\frac{-dx}{x}$

Integrate on both sides,

$\large \int \frac{dy}{y}= \int \frac{-dx}{x}$

$=>\log y=-\log x+\log c$

$=>\log y +\log x=\log c$

$=>\log (xy)=\log c$

or $xy=c$

It is given that the curve passes through the point $(1,1)$.Hence substituting $x=1$ and $y=1$

$1 \times 1=c=>c=1$

Therefore the required solution is $xy=1$