jmc

To find the standard form for the equation of the hyperbola, we complete the square in both variables: $$\begin{array}{rl}-(x^2+10x)+2(y^2-8y)+1& =0\\ -(x^2+10x+25)+2(y^2-8y+16)+1& =-25+32\\ -(x+5{)}^2+2(y-4{)}^2& =6\\ \frac{(y-4{)}^2}{3}-\frac{(x+5{)}^2}{6}& =1.\end{array}$$This fits the standard form of the hyperbola $$\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1,$$where $a=\sqrt{3},$ $b=\sqrt{6},$ $h=-5,$ and $k=4.$ Thus, the center of the hyperbola is the point $(h,k)=(-5,4).$ Because the $y^2$ coefficient is positive and the $x^2$ coefficient is negative, the foci are vertically aligned with the center of the hyperbola. We have $$c=\sqrt{a^2+b^2}=\sqrt{3+6}=3,$$which is the distance from the center of the hyperbola to each focus. Therefore, the two foci of the hyperbola are $(-5,4\pm 3),$ which gives two points: $\boxed{(-5,7)}$ and $\boxed{(-5,1)}.$ (Either point is an acceptable answer.)