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

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(C) The ellipse $E_1$ 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

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$1/2$

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(C) The ellipse $E_1$ 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 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 $E_2$ circumscribes the rectangle $R$,the points $(\pm 3, \pm 2)$ lie on $E_2$. Thus,$\frac{9}{a^2} + \frac{4}{b^2} = 1$.Given that $E_2$ passes through $(0, 4)$,we have $b = 4$,so $b^2 = 16$.Substituting $b^2 = 16$ into the equation: $\frac{9}{a^2} + \frac{4}{16} = 1 \Rightarrow \frac{9}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}$.Therefore,$a^2 = \frac{9 	imes 4}{3} = 12$.For an ellipse with $b > a$,the eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$.$12 = 16(1 - e^2) \Rightarrow 1 - e^2 = \frac{12}{16} = \frac{3}{4}$.$e^2 = 1 - \frac{3}{4} = \frac{1}{4}$.$e = \frac{1}{2}$.

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 $E_2$ circumscribes the rectangle $R$ and passes through the point $(0, 4)$. What is the eccentricity of the ellipse $E_2$?

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