2. Given the curve $y^{2}=a x$ and its curve symmetric about the point $(1,1)$ intersect at two different points. If the inclination angle of the line passing through these two points is $45^{\circ}$, then the value of the real number $a$ is $(\quad)$.
(A)2
(B) 4
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$
2. (A).

The curve $y^{2}=a x$ symmetric to the point $(1,1)$ is $(2-y)^{2}=a(2-x)$.
From $\left\{\begin{array}{l}y^{2}=a x, \\ (2-y)^{2}=a(2-x)\end{array}\right.$ we get $y^{2}-2 y+2-a=0$.
$\therefore y_{1}+y_{2}=2$.
Thus, $k=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{a\left(y_{1}-y_{2}\right)}{y_{1}^{2}-y_{2}^{2}}=\frac{a}{y_{1}+y_{2}}=\frac{a}{2}=1$.
$\therefore a=2$