A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, then P is the mid-point of AB. The curve passes through the point (1, 1)  is  xy=c² then c² =-----
 

Equation of the tangent at P(x, y) to the curve   

$\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\underset{\_}{\text{ is }}\mathrm{Y}-\mathrm{y}=\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{X}-\mathrm{x})$
Putting $\underset{\_}{\mathrm{Y}}=0;\mathrm{X}=\mathrm{X}-\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}$
and for $\mathrm{X}=\underset{\_}{\underset{\_}{0,\mathrm{Y}}}=\mathrm{y}-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \mathrm{A}\equiv \left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{dy}/\mathrm{dx}},0\right),\mathrm{B}\equiv \left(0,\mathrm{y}-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right)\text{. }$
Since P is the mid-point of AB, $\mathrm{x}=\frac{\mathrm{x}-\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}}{2}\text{ and }\mathrm{y}=\frac{\mathrm{y}-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}}{2}$
$\text{ or }\mathrm{x}=-\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}\text{ and }\mathrm{y}=-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\text{. }$
Both of these equations give  $\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \frac{\mathrm{dy}}{\mathrm{y}}+\mathrm{c}\underset{\_}{\underset{\_}{\text{ or }\ln }\mathrm{x}}=-\ln \mathrm{y}+\ln \mathrm{c}\text{ or }\mathrm{x}=\mathrm{c}/\mathrm{y}$
As the curve passes through (1, 1), c = 1. 
Equation of the curve is xy = 1.