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The foci of the ellipse $9 x^2+25 y^2=225$ are
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Verified Answer
The correct answer is:
$( \pm 4,0)$
Ellipse is $9 x^2+25 y^2=225$
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1, a=5, b=3 \\
& \therefore e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}
\end{aligned}
$$
$\therefore$ Foci are $( \pm a e, 0) \equiv( \pm 4,0)$
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1, a=5, b=3 \\
& \therefore e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}
\end{aligned}
$$
$\therefore$ Foci are $( \pm a e, 0) \equiv( \pm 4,0)$
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