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 $$AB$$.

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 $AB$.
So, the coordinates of points $A$ and $B$

are $(2x,0)$ and $(0,2y)$, respectively.
$\therefore$ Slope of $AB = \frac{{0 - 2y}}{{2x - 0}} = - \frac{y}{x}$

Since, the segment $AB$
is a tangent to the curve at $P$.

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

$\Rightarrow$ $\quad \frac{{dy}}{y} = - \frac{{dx}}{x}$

On integrating both sides,

we get $\log y = - \log x + \log C$
$\log y = \log \frac{C}{x}$

….(i)
Since, the given curve passes through (1,1)

$\therefore$ $\log 1 = \log \frac{C}{1}$

$\Rightarrow$ $0 = \log C$

$\Rightarrow$ $c = 1$

$\therefore \log y = \log \frac{1}{x}$

$\Rightarrow$ $y = \frac{1}{x}$

$\Rightarrow$ $xy = 1$