If the line $$\alpha x+4y=\sqrt{7}$$, where $$\alpha \epsilon R$$, touch the ellipse $$3x^{2}+4y^{2}=1$$ at the point P in the first quadrant, then one of the focal distances of P is:

A

$$\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{11}}$$

B

$$\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$$

C

$$\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}$$

D

$$\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{5}}$$