Equation of the Tangent to the Curve

$(y-2)^{2} = x \\ y^2-4y+4=x\\ \frac{d}{dx}(y^2-4y+4) = \frac{d}{dx}(x)\\ 2yy'-4y'=1\\y'(2y-4)=1\\y'=\frac{1}{2y-4}=\frac{1}{2}\\ 2=2y-4 \Longrightarrow y=3\\------------\\(y-2)^2=x \Longrightarrow (3-2)^2=x \Longrightarrow x= 1\\\\$

Now we have found the point on y where the slope is 1/2 (slope needs to be 1/2 in order to be parallel to x-2y=4). We can now make an equation with point slope form.

$y-3 = \frac{1}{2}(x-1) \Longrightarrow1x-2y+5=0\\ A = 1\\B = 2\\C=5\\A-B+C=1-2+5=\boxed{4}$

$(y-2)^2=x$

$\Rightarrow 2(y-2)\frac {dy}{dx} = 1 \quad \Rightarrow \frac{dy}{dx} = \frac {1} {2(y-2)}$

$x - 2y = 4\quad \Rightarrow y = \frac {1}{2}x -4$

Therefore gradient of the tangent = $\frac {1}{2} = \frac {dy} {x} = \frac {1} {2(y-2)}$

At the tangent, $y - 2 = 1\quad \Rightarrow y = 3$ and $x = (3-2)^2 = 1$

The tangent line is given by: $y = \frac {1}{2} + c\quad \Rightarrow 3 = \frac{1}{2} + c \quad \Rightarrow c = \frac{5}{2}$

Therefore, $y = \frac {1}{2}x + \frac{5}{2} \quad \Rightarrow x - 2y + 5 = 0 \quad \Rightarrow A - B + C = 1 - 2 + 5 = \boxed {4}$

Given equation (y-2)^2 = x...Differentiating both sides, we get 2(y-2) dy/dx =1...now since the tangent is parallel to x-2y=4 or y=(1/2)x +2... so slope of the tangent = slope of this line...therefore dy/dx = 1/2... putting this in above we get 2 (y-2) (1/2)=1 so y=3...therefore x=1....therefore equation of tangent is (y-3)=(1/2)*(x-1) so x-2y+5=0...So A=1 B=2 C=5...