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In an ellipse $9 x^2+5 y^2=45$, the distance between the foci is
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$4$
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 $\left.b^2 \gt a^2\right) a^2=b^2\left(1-e^2\right) \lambda$
or $e^2=\frac{b^2-a^2}{b^2}=\frac{9-5}{9}=\frac{4}{9}$ or $e=\frac{2}{3}$.
Therefore distance between foci $=2 b e=2 \times 3 \times \frac{2}{3}=4$.
or $e^2=\frac{b^2-a^2}{b^2}=\frac{9-5}{9}=\frac{4}{9}$ or $e=\frac{2}{3}$.
Therefore distance between foci $=2 b e=2 \times 3 \times \frac{2}{3}=4$.
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