Now assuming that the curve is y = f(x), we get the equation of the tangent at any point (x,y) on curve as

                          Y - y = dy/dx ( X - x)

 So the point at which tangent meets X axis is ( (x(dy/dx) - y)/(dy/dx)) , 0)

 and the point at which it meets Y axis is ( 0 , y - (dy/dx.)x )

So the area of triangle formed by the tangent and the co-ordinate axis is

 =  1/2 * ((x(dy/dx) - y)/(dy/dx))*(y - (dy/dx.)x) = 2 (given)

 So solving this equation we get a quadratic in dy/dx :

              x² (dy/dx)² - (2xy - 4)dy/dx + y² = 0

   Solving we get dy/dx = (2xy - 4 + 4(xy - 2)² - 4(xy)²) / 2 x²

             &          dy/dx = (2xy - 4 - 4(xy - 2)² - 4(xy)²) / 2 x²

    Solve these equations by putting xy = t and then using the boundary condition i.e the curve passes through (1,1).

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