If e and e^{prime} are the eccentricities of | vedclass

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(A) The equation of the ellipse is $5x^2 + 9y^2 = 45$,which can be written as $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here,$a^2 = 9$ and $b^2 = 5$. The e

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(A) The equation of the ellipse is $5x^2 + 9y^2 = 45$,which can be written as $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here,$a^2 = 9$ and $b^2 = 5$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$. The equation of the hyperbola is $5x^2 - 4y^2 = 45$,which can be written as $\frac{x^2}{9} - \frac{y^2}{45/4} = 1$. Here,$a^2 = 9$ and $b^2 = \frac{45}{4}$. The eccentricity $e^{\prime}$ is given by $e^{\prime} = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{45/4}{9}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$. Therefore,$ee^{\prime} = \frac{2}{3} 	imes \frac{3}{2} = 1$.

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

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