Given that the tangent of the curve y = ln (x-2a) + √ (1 + ax) at x = 0 is parallel to the X axis. 1. Find the value of A. 2. Find the tangent and normal equation of the curve at x = 0

Given that the tangent of the curve y = ln (x-2a) + √ (1 + ax) at x = 0 is parallel to the X axis. 1. Find the value of A. 2. Find the tangent and normal equation of the curve at x = 0

(1) When x = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 0, y ′ = 1 / (2a) = A / 2, a / 2, a / 2, a / 2, a / 2, a / 2, a / 2, and a = 1, y = (X-2) (X-2) + (1 + x) at this time, the domain does not contain x = 0. Rounding. When a = -1, when a = -1, y = (1, y = (x (x + 2) y = (2) y = (13265㏑ 2 ㏑ 2 ㏑ 2 ㏑ 2 y '(0) = 0. The tangent equation is y = 1 + ㏑ 2. The normal equation is x = 0