The ellipse E_1: frac{x^2}{9}+frac{y^2}{4}=1 | vedclass

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(C) The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$.Since $E_1$ is inscribed in a rectangle $R$ with si

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$\frac{1}{2}$

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(C) The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$.Since $E_1$ is inscribed in a rectangle $R$ with sides parallel to the coordinate axes,the vertices of the rectangle $R$ are $(\pm 3, \pm 2)$.Let the equation of the ellipse $E_2$ be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.Since $E_2$ passes through $(0, 4)$,we have $\frac{0}{a^2}+\frac{16}{b^2}=1$,which gives $b^2=16$.Since $E_2$ circumscribes the rectangle $R$,it passes through the point $(3, 2)$.Substituting $(3, 2)$ into the equation of $E_2$: $\frac{9}{a^2}+\frac{4}{b^2}=1$.Substituting $b^2=16$: $\frac{9}{a^2}+\frac{4}{16}=1$ $\Rightarrow \frac{9}{a^2}+\frac{1}{4}=1$ $\Rightarrow \frac{9}{a^2}=\frac{3}{4}$ $\Rightarrow a^2=12$.For the ellipse $E_2$,$b^2 > a^2$ (since $16 > 12$),so the eccentricity $e$ is given by $a^2=b^2(1-e^2)$.$12=16(1-e^2) \Rightarrow 1-e^2=\frac{12}{16}=\frac{3}{4}$.$e^2=1-\frac{3}{4}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$.

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 the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_2$ is

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