Ellipse

Concept:For a line to be tangent to an ellipse, the condition $c^2 = a^2 m^2 + b^2$ must hold.Explanation:The given line is $\alpha x + 4y = \sqrt{7}$.Rewrite it as $y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$.So slope $m = -\frac{\alpha}{4}$ and intercept $c = \frac{\sqrt{7}}{4}$.The ellipse is $3x^2 + 4y^2 = 1$, i.e., $\frac{x^2}{\frac{1}{3}} + \frac{y^2}{\frac{1}{4}} = 1$.Thus $a^2 = \frac{1}{3}$, $b^2 = \frac{1}{4}$.Tangency condition: $c^2 = a^2 m^2 + b^2$ gives $\frac{7}{16} = \frac{1}{3} \cdot \frac{\alpha^2}{16} + \frac{1}{4}$.Solve: $\frac{\alpha^2}{48} = \frac{7}{16} - \frac{4}{16} = \frac{3}{16}$ $\Rightarrow$ $\alpha^2 = 9$ $\Rightarrow$ $\alpha = \pm 3$.Since point P is in the first quadrant, we take $\alpha = 3$.Tangent line: $3x + 4y = \sqrt{7}$.Point of contact P$(x_1, y_1)$ satisfies $\frac{3x_1}{3} = \frac{4y_1}{4} = \frac{1}{\sqrt{7}}$.Thus $x_1 = \frac{1}{\sqrt{7}}$, $y_1 = \frac{1}{\sqrt{7}}$.Eccentricity: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/4}{1/3}} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$.Focal distances: $PS = a - e x_1$ and $PS' = a + e x_1$ (foci at $(\pm ae, 0)$).Here $a = \frac{1}{\sqrt{3}}$, $e = \frac{1}{2}$, $x_1 = \frac{1}{\sqrt{7}}$.So $PS' = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$ and $PS = \frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{7}}$.Among the options, $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$ is present.Answer:Option D: $\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$