Let the curve C be the mirror image of the parabola $y^{2} = 4x$ with respect to the line $x + y + 4 = 0.$ If A and B are the points of intersection of C with the line $y = -5,$ then the distance between A and B is

$\text{Given: parabola } y^2 = 4x$ $\text{Mirror line: } x + y + 4 = 0$ $C = \text{mirror of } y^2 = 4x \text{ in } x + y + 4 = 0$ $\text{To find: } AB, \text{ where } A, B \text{ lie on } C \text{ and } y = -5$ $x + y + 4 = 0 \Rightarrow y = -x - 4$ $x' = x - \dfrac{2a(ax+by+c)}{a^2+b^2}$ $y' = y - \dfrac{2b(ax+by+c)}{a^2+b^2}$ $a=b=1,\ c=4 \Rightarrow a^2+b^2=2$ $x' = x-(x+y+4) = -y-4$ $y' = y-(x+y+4) = -x-4$ $X = -y-4,\quad Y = -x-4$ $\Rightarrow y = -X-4,\quad x = -Y-4$ $(-X-4)^2 = 4(-Y-4)$ $(X+4)^2 = -4Y - 16 = -4(Y+4)$ $C:\ (x+4)^2 = -4(y+4)$ $\text{Vertex of } C:\ (-4,-4)$ $(x+4)^2 = -4(-5+4) = 4$ $x+4 = \pm 2 \Rightarrow x = -2\ \text{or}\ -6$ $A(-2,-5),\quad B(-6,-5)$ $AB = |-2-(-6)| = 4$ $\text{mirror of } y=-5 \text{ is } x=1$ $y^2 = 4(1) = 4 \Rightarrow y = \pm 2$ $AB = 2-(-2) = 4$ $\boxed{AB = 4}$