Let e be the eccentricity of the ellipse frac | vedclass

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(D) Given the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$. Since $b^2 > a^2$ $(9 > 4)$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} =

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$\frac{3}{2}$

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(D) Given the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$. Since $b^2 > a^2$ $(9 > 4)$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. Let $e_H$ be the eccentricity of the hyperbola,where $e_H = \frac{1}{e} = \frac{3}{\sqrt{5}}$. For a hyperbola with eccentricity $e_H$ and its conjugate hyperbola with eccentricity $e_C$,the relation is $\frac{1}{e_H^2} + \frac{1}{e_C^2} = 1$. Substituting $e_H = \frac{3}{\sqrt{5}}$,we get $\frac{1}{(3/\sqrt{5})^2} + \frac{1}{e_C^2} = 1$. $\frac{5}{9} + \frac{1}{e_C^2} = 1 \Rightarrow \frac{1}{e_C^2} = 1 - \frac{5}{9} = \frac{4}{9}$. Therefore,$e_C^2 = \frac{9}{4}$,which gives $e_C = \frac{3}{2}$.

Let $e$ be the eccentricity of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$. If $\frac{1}{e}$ is the eccentricity of a hyperbola,then the eccentricity of its conjugate hyperbola is

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