In an ellipse 9x^2 + 5y^2 = 45,the distance b | vedclass

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(D) The given equation is $9x^2 + 5y^2 = 45$.Dividing both sides by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.Comparing this with the standard f

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(D) The given equation is $9x^2 + 5y^2 = 45$.Dividing both sides by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 5$ and $b^2 = 9$.Since $b^2 > a^2$,the major axis is along the $y$-axis.The eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$.Thus,$e = \frac{2}{3}$.The distance between the foci is $2be = 2 	imes \sqrt{9} 	imes \frac{2}{3} = 2 	imes 3 	imes \frac{2}{3} = 4$.

In an ellipse $9x^2 + 5y^2 = 45$,the distance between the foci is

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