Let Γ denote a curve y=f(x) which is in the first quadrant and let the point (1,0) lie on it. Let the tangent to Γ at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on Γ. Then which of the following options is/are correct?

Correct option is D.
y=ln(1+√(1-x²/x))-1-√(1-x²)
Let (a,f(a)) be a point on the curve. Let f’(x) be the differentiation of f(x).
Equation of tangent (y-f(a)) = f’(a)(x-a)
Put x=0
y-f(a) = -af’(a)
y = f(a) - af’(a)
YP = (0, f(a) - af’(a))
P = (a, f(a))
PYP = √(a² + (af’(a))²) = 1
a² + a²(f’(a))² = 1
(f’(x))² = (1-x²)/x²
∫f’(x)dx = ±∫√(1-x²)/x dx
Put 1-x² = t² => 2xdx = -2tdt => xdx = -tdt
∫f’(x)dx = ±∫-t²/√(1-t²)dt = ±∫(1-(1-t²))/√(1-t²) dt = ±∫(1/√(1-t²)-√(1-t²))dt
= ±(sin⁻¹t - ∫√(1-t²)dt)
= ±(sin⁻¹t + (t/2)√(1-t²) + (1/2)sin⁻¹t) + c
= ±((3/2)sin⁻¹t + (t/2)√(1-t²)) + c
= ±((3/2)sin⁻¹√(1-x²) + ((√(1-x²))/2)x) + c
Put x=1 and y=0 => c=0