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 AB.

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 $xy=1$