The ellipse $E_1 : \frac{x^2}{9} + \frac{y^2}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through $(0, 4)$ circumscribes the rectangle $R$. The eccentricity of $E_2$ is:
Step-by-Step Solution
Key Concept: $R$ has corners $(\pm 3, \pm 2)$. Let $E_2 : x^2/A^2 + y^2/B^2 = 1$. Passes through $(0, 4)$: $B^2 = 16$. Passes through $(3, 2)$: $9/A^2 + 4/16 = 1 \Rightarrow A^2 = 12$. Since $B > A$, $e^2 = 1 - A^2/B^2 = 1/4$, so $e = 1/2$.
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Correct Answer: (3)