If the line $$\alpha x + 2y = 1$$, where $$\alpha \in R $$, does not meet the hyperbola $$x^{2}-9y^{2}=9$$, then a possible value of $$\alpha$$ is:

Standard form of Hyperbola: $$\frac{x^2}{9} - \frac{y^2}{1} = 1$$ (where $$a=3, b=1$$).

Condition for non-intersection:

A line $$y = mx + c$$ does not meet a hyperbola if it passes between the two branches or doesn't touch them. Specifically, for the line to not intersect, $$c^2 > a^2m^2 - b^2$$ must be false (it must intersect) OR we look at the slopes.

Rewrite the line: $$2y = -\alpha x + 1 \implies y = (-\frac{\alpha}{2})x + \frac{1}{2}$$.

Here $$m = -\frac{\alpha}{2}$$ and $$c = \frac{1}{2}$$.

The condition for a line to be a tangent is $$c^2 = a^2m^2 - b^2$$.

For it to not meet, we need $$c^2 < a^2m^2 - b^2$$.

$$(\frac{1}{2})^2 < (3^2)(-\frac{\alpha}{2})^2 - 1^2$$

$$\frac{1}{4} < 9(\frac{\alpha^2}{4}) - 1 \implies \frac{5}{4} < \frac{9\alpha^2}{4} \implies \alpha^2 > \frac{5}{9} \approx 0.555$$

Evaluate options:

o A: $$0.6^2 = 0.36$$ (False)

o B: $$0.5^2 = 0.25$$ (False)

o C: $$0.7^2 = 0.49$$ (False)

o D: $$0.8^2 = 0.64$$ (True, as $$0.64 > 0.555$$)

Correct Option: D