The two conics _{$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$} and _{$y^2=-\frac{b}{a}x$} intersect iff

The x co-ordinates of the points of intersection are given by _{$\frac{x^2}{a^2}+\frac{1}{ab}x+1=0$}

And the roots are real iff _{$\frac{1}{a^2{b}^2}-\frac{4}{a^2}\ge 0\Rightarrow \frac{1}{b^2}-4\ge 0$}

_{$\Rightarrow 0<b^2\le \frac{1}{4}\Rightarrow 0<b\le \frac{1}{2}$}