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

Let $y^{2}=a x$ be the curve symmetric to point $(1,1)$. For any point $(x, y)$ on this curve, $\left(x^{\prime}, y^{\prime}\right)$ is the symmetric point of $(x, y)$ about point $(1,1)$. Then,
$$
y^{\prime 2}=a x^{\prime} \text {. }
$$

Also, $x+x^{\prime}=2$ and $y+y^{\prime}=2$, so $x^{\prime}=2-x$ and $y^{\prime}=2-y$. Substituting into equation (1) gives
$$
(2-y)^{2}=a(2-x) \text {. }
$$

Equation (2) is the equation of the curve $y^{2}=a x$ symmetric to point $(1,1)$. Let the intersection points be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$. Then these two points are also symmetric about $(1,1)$. Therefore,
$$
\begin{array}{l}
y_{1}+y_{2}=2, \\
\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\tan 45^{\circ}=1 . \\
\text { Also, } y_{1}^{2}=a x_{1}, y_{2}^{2}=a x_{2}, \text { so } \\
y_{1}^{2}-y_{2}^{2}=a\left(x_{1}-x_{2}\right),
\end{array}
$$

which means $\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)=a\left(x_{1}-x_{2}\right)$.
Thus, $a=\left(y_{1}+y_{2}\right) \cdot \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2$.