Ellipse

Concept:The line $y = mx + c$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if and only if $c^2 = a^2 m^2 + b^2$.For the tangent $lx + my = n$, the point of contact is $\left(\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)$.The focal distances of a point $P(x_1, y_1)$ on the ellipse are $a - ex_1$ and $a + ex_1$.Explanation:The ellipse is $3x^2 + 4y^2 = 1$, i.e. $\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$, so $a^2 = \frac{1}{3}$, $b^2 = \frac{1}{4}$, with the major axis along the $x$-axis.The given line $\alpha x + 4y = \sqrt{7}$ can be written as $y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$, so $m = -\frac{\alpha}{4}$ and $c = \frac{\sqrt{7}}{4}$.Applying the tangency condition $c^2 = a^2 m^2 + b^2$:$\frac{7}{16} = \frac{1}{3}\cdot\frac{\alpha^2}{16} + \frac{1}{4}$.Thus $\frac{\alpha^2}{48} = \frac{7}{16} - \frac{4}{16} = \frac{3}{16}$, giving $\alpha^2 = 9$, i.e. $\alpha = \pm 3$.Since $P$ lies in the first quadrant, we take $\alpha = 3$ (for $\alpha = -3$ the point of contact lies in the second quadrant).For the tangent $3x + 4y = \sqrt{7}$, the point of contact is $P\left(\frac{(1/3)\cdot 3}{\sqrt{7}}, \frac{(1/4)\cdot 4}{\sqrt{7}}\right) = \left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$.Eccentricity: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/4}{1/3}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.So $ae = \frac{1}{2\sqrt{3}}$ and the foci are $\left(\pm\frac{1}{2\sqrt{3}}, 0\right)$.Hence the focal distances of $P\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$ are:$a - ex_1 = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{7}}$ and $a + ex_1 = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.Among the given options, only $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$ is present.Answer:Option D: $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$