The point on the curve $4 y^2-4 y+2 x-1=0$ at which the tangent becomes parallel toY-axis is

The condition for a tangent to be parallel to the Y-axis is that $\frac{dx}{dy} = 0$.
Differentiate $4y^2 - 4y + 2x - 1 = 0$ with respect to $y$:

$8y - 4 + 2\frac{dx}{dy} = 0$

Solving for $\frac{dx}{dy}$ gives $\frac{dx}{dy} = 2 - 4y$.

Set $\frac{dx}{dy} = 0$, so $2 - 4y = 0$, yielding $y = \frac{1}{2}$.

Substitute into the original equation:

$4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 2x - 1 = 0$

$1 - 2 + 2x - 1 = 0$

$2x - 2 = 0$, so $x = 1$.

The point is $\left(1, \frac{1}{2}\right)$, corresponding to option A.