From the line: $y=\frac{1-\alpha x}{2}$. Substitute into the hyperbola:

$x^2-9{\left(\frac{1-\alpha x}{2}\right)}^2=9\Rightarrow 4x^2-9(1-\alpha x{)}^2=36\Rightarrow (4-9{\alpha }^2)x^2+18\alpha x-45=0$.

For no real intersection, this quadratic in $x$ must have negative discriminant:

$D=(18\alpha {)}^2+4\cdot 45\cdot (4-9{\alpha }^2)=324{\alpha }^2+720-1620{\alpha }^2=720-1296{\alpha }^2<0$.

So ${\alpha }^2>\frac{720}{1296}=\frac{5}{9}$, i.e. $|\alpha |>\frac{\sqrt{5}}{3}\approx 0.745$.

Among the options, only $0.8>0.745$. So $\alpha =0.8$.