A curve passes through (1, 1) such that triangle formed by coordinate axes and tangent at any point of curve is in first quadrant and has its area equal to 2. What will be equation of curve?

The equation of tangent to the curve y = f(x), at point (x, y), is
Y ? y = dy/dx * (X ? x) ?..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))² ? 4dy/dx = 0
Or, x²(dy/dx)² ? 2(xy ? 2)dy/dx + y² = 0
Solving for dy/dx we get,
dy/dx = [(xy ? 2) ± √(1 ? xy)]/ x²
Let, 1 ? xy = t²
=> x(dy/dx) + y = -2t(dt/dx)
=> x²(dy/dx) = t² ? 1 ? 2tx(dt/dx), so that (3) gives
t(x(dt/dx) ? (t ± 1)) = 0
Hence, either t = 0
=>xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
=> dx/x = dt/t ± 1
=> t ± 1 = cx
For x = 1, y = 1 and t = 0
=> c = ± 1, so the solution is
t = ± (x ? 1) => t² = (x ? 1)²
Or, 1 ? xy = x² ? 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2