If the line alpha x+4y=sqrt{7}, where alpha i | vedclass

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(D) The equation of the line is $\alpha x+4y-\sqrt{7}=0$. This line touches the ellipse $3x^{2}+4y^{2}=1$, which can be written as $\frac{x^{2}}{1/3}

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

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(D) The equation of the line is $\alpha x+4y-\sqrt{7}=0$. This line touches the ellipse $3x^{2}+4y^{2}=1$, which can be written as $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$. Comparing with the condition of tangency $c^{2}=a^{2}m^{2}+b^{2}$, where $m = -\frac{\alpha}{4}$ and $c = \frac{\sqrt{7}}{4}$:$(\frac{\sqrt{7}}{4})^{2} = (\frac{1}{3})(\frac{-\alpha}{4})^{2} + \frac{1}{4} \implies \frac{7}{16} = \frac{\alpha^{2}}{48} + \frac{1}{4} \implies \frac{\alpha^{2}}{48} = \frac{3}{16} \implies \alpha^{2} = 9 \implies \alpha = \pm 3$. Since $P$ is in the first quadrant, the tangent is $3x+4y=\sqrt{7}$. The point of contact $P(x_{1}, y_{1})$ for the tangent $3x+4y=\sqrt{7}$ is given by $\frac{3x}{x_{1}} = \frac{4y}{y_{1}} = \frac{1}{1/7}$ is not correct, rather using $3xx_{1}+4yy_{1}=1$, we get $3x_{1} = \frac{3}{\sqrt{7}}$ and $4y_{1} = \frac{4}{\sqrt{7}}$, so $P = (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$. For the ellipse $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$, $a^{2} = 1/3$ and $b^{2} = 1/4$. Eccentricity $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1/4}{1/3}} = \sqrt{1-\frac{3}{4}} = \frac{1}{2}$. The focal distances are $a \pm ex_{1} = \frac{1}{\sqrt{3}} \pm \frac{1}{2}(\frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$. Thus, one of the focal distances is $\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.

If the line $\alpha x+4y=\sqrt{7}$, where $\alpha \in R$, 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:

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