If the line \(\alpha x+4y=\sqrt{7} \), where \( a\in \mathbb{R} \), touches the ellipse \( 3x^2+4y^2=1 \) at the point \( P \) in the first quadrant, then one of the focal distances of \( P \) is:
\( \frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{11}} \)
\( \frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}} \)
\( \frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{5}} \)
\( \frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{5}} \)