9. Does there exist a translation such that the parabola $y=-x^{2}$, after translation, passes through the origin and the triangle formed by the vertex of the translated parabola and its two intersections with the $x$-axis has an area of 1? If it exists, find the equation of the function; if not, explain the reason.
9. Suppose there exists such a translation $a=(h, k)$. From $\left\{\begin{array}{l}x=x^{\prime}-h \\ y=y^{\prime}-k\end{array}\right.$, substituting into $y=-x^{2}$ yields $y^{\prime}-k=-\left(x^{\prime}-h\right)^{2}$, which means the translated parabola is $y=-(x-h)^{2}+k$, with the vertex at $(h, k)$. According to the given condition, this curve passes through the origin, so $k=h^{2}$. Setting $y=0$, we get $x=h \pm \sqrt{k}$, so the length of the segment it intercepts on the $x$-axis is $2 \sqrt{k}$. By the problem, $\frac{1}{2} \cdot 2 \sqrt{k} \cdot k=1$, solving this gives $k=1, h= \pm 1$. Therefore, such a translation $\boldsymbol{a}=(1,1)$ or $(-1,1)$ exists. When $\boldsymbol{a}=(1,1)$, the translated equation is $y=-(x-1)^{2}+1$; when $\boldsymbol{a}=$ $(-1,1)$, the translated equation is $y=-(x+1)^{2}+1$.