jmc

To try and write the given equation in standard form, we complete the square in each variable: $$\begin{array}{rl}(x^2-10x)+4(y^2+14y)& =k\\ (x^2-10x+25)+4(y^2+14y+49)& =k+25+4(49)=k+221\\ (x-5{)}^2+4(y+7{)}^2& =k+221.\end{array}$$We see that if $k+221>0,$ then we can divide both sides by $k+221$ to obtain the standard form for the equation of an ellipse. On the other hand, if $k+221=0,$ then this equation is only satisfied when $x-5=0$ and $y+7=0,$ so the graph of the equation only consists of a single point. And if $k+221<0,$ then no points $(x,y)$ satisfy this equation. Therefore, the graph is a non-degenerate ellipse if and only if $k+221>0,$ that is, $k>-221.$ Thus, $a=\boxed{-221}.$