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:

For ellipse ($$3x^2+4y^2=1):$$

$$a^2=\frac{1}{3},\quad b^2=\frac{1}{4}\Rightarrow a=\frac{1}{\sqrt{3}},;c^2=a^2-b^2=\frac{1}{12}\Rightarrow c=\frac{1}{2\sqrt{3}},;e=\frac{c}{a}=\frac{1}{2}$$

Tangent condition give$$s(\alpha^2=9\Rightarrow\alpha=3)$$ (first quadrant).

Point of contact:
$$x=\frac{\alpha\sqrt{7}}{12+\alpha^2}=\frac{\sqrt{7}}{7},\quad y=\frac{\sqrt{7}}{7}$$

Focal distances for ellipse:
$$r_1,r_2=a\pm ex$$
=$$\frac{1}{\sqrt{3}}\pm\frac{1}{2}\cdot\frac{\sqrt{7}}{7}$$
=$$\frac{1}{\sqrt{3}}\pm\frac{1}{2\sqrt{7}}$$