13) A

y = ⎡x² + 2ax⎤½
       ⎣     b       ⎦

y² = x² + 2ax
            b    

by² = x² + 2ax

This is a bit tricky. (x² + 2ax) is part of (x + a) (x + a) = x² + 2ax + a²

(x + a) (x + a) = x² + 2ax + a²
x² + 2ax = (x + a) (x + a) – a²

So we get,

by² = (x + a) (x + a) – a²
by² + a² = (x + a)²

Rooting both LHS and RHS, 

(by² + a²)^½ = x + a

x = (by² + a²)^½ – a

So the answer is A.