In an ellipse 9{x^2} + 5{y^2} = 45, the dista | vedclass

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(d) Given equation may be written as $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{9} = 1.$

Comparing the given equation with standard equation, we get ${a^2

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$4$

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(d) Given equation may be written as $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{9} = 1.$

Comparing the given equation with standard equation, we get ${a^2} = 5$ and ${b^2} = 9.$

We also know that in an ellipse (where ${b^2} > {a^2})$ ${a^2} = {b^2}(1 - {e^2})\,\lambda $ or ${e^2} = \frac{{{b^2} - {a^2}}}{{{b^2}}} = \frac{{9 - 5}}{9} = \frac{4}{9}$

$e = \frac{2}{3}.$

Therefore distance between foci $ = 2be = 2 	imes 3 	imes \frac{2}{3} = 4$.

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

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