Let E_1 = frac{x^2}{9} + frac{y^2}{4} = 1 and | vedclass

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(D) For the inscribed ellipse $E_1: \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,the rectangle $R$ has vertices at $(\pm 3, \pm 2)$.Since $E_2: \frac{x^2}{a

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$a = 2\sqrt{3}, b = 4$

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(D) For the inscribed ellipse $E_1: \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,the rectangle $R$ has vertices at $(\pm 3, \pm 2)$.Since $E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ circumscribes the rectangle $R$,it must pass through the vertices of $R$,such as $(3, 2)$.Substituting $(3, 2)$ into the equation of $E_2$: $\frac{3^2}{a^2} + \frac{2^2}{b^2} = 1 \Rightarrow \frac{9}{a^2} + \frac{4}{b^2} = 1$.Given that $E_2$ passes through $(0, 4)$,we have $\frac{0^2}{a^2} + \frac{4^2}{b^2} = 1$,which gives $b^2 = 16$,so $b = 4$.Substituting $b^2 = 16$ into the equation $\frac{9}{a^2} + \frac{4}{16} = 1$:$\frac{9}{a^2} + \frac{1}{4} = 1$ $\Rightarrow \frac{9}{a^2} = \frac{3}{4}$ $\Rightarrow a^2 = 12$ $\Rightarrow a = 2\sqrt{3}$.Thus,$a = 2\sqrt{3}$ and $b = 4$.

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