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The foci of the hyperbola $9 x^2-16 y^2=144$ are
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The correct answer is:
$( \pm 5,0)$
The equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$
Now $b^2=a^2\left(e^2-1\right) \Rightarrow e=\frac{5}{4}$
Hence foci are $( \pm a e, 0) \Rightarrow\left( \pm 4 \cdot \frac{5}{4}, 0\right)$ i.e., $( \pm 5,0)$
Now $b^2=a^2\left(e^2-1\right) \Rightarrow e=\frac{5}{4}$
Hence foci are $( \pm a e, 0) \Rightarrow\left( \pm 4 \cdot \frac{5}{4}, 0\right)$ i.e., $( \pm 5,0)$
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