Eccentricity of Circumscribing Ellipse Around Bounding Rectangle

To find the eccentricity of the ellipse $E_2$, we proceed as follows:

1. Determine the vertices of the rectangle $R$: The given ellipse is: $E_1 : \frac{x^2}{9} + \frac{y^2}{4} = 1$ Since the rectangle $R$ circumscribes $E_1$ and has sides parallel to the coordinate axes, the sides of the rectangle are tangent to $E_1$ at its vertices: $x = \pm 3 \quad \text{and} \quad y = \pm 2$ Therefore, the four vertices of the rectangle $R$ are: $(\pm 3, \pm 2)$

2. Formulate the equation of the ellipse $E_2$: The ellipse $E_2$ circumscribes the rectangle $R$, which is symmetric with respect to both the coordinate axes and centered at the origin $(0,0)$. Thus, the equation of $E_2$ can be assumed to be of the standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

3. Find the parameters $a^2$ and $b^2$:

   • Since $E_2$ passes through the point $(0, 4)$, substituting this point into the equation gives: $\frac{0^2}{a^2} + \frac{4^2}{b^2} = 1 \implies b^2 = 16$
   • Since $E_2$ circumscribes the rectangle $R$, it passes through the vertex $(3, 2)$. Substituting $(3, 2)$ and $b^2 = 16$ into the equation: $\frac{3^2}{a^2} + \frac{2^2}{16} = 1$ $\frac{9}{a^2} + \frac{1}{4} = 1$ $\frac{9}{a^2} = \frac{3}{4} \implies a^2 = 12$

4. Calculate the eccentricity of $E_2$: Notice that $b^2 = 16 > a^2 = 12$, which indicates that the major axis of $E_2$ lies along the $y$-axis. The eccentricity $e$ is given by: $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Thus, the eccentricity of the ellipse $E_2$ is $\frac{1}{2}$, which corresponds to Option C.