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If $e_{1}$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $e_{2}$ is the eccentricity
of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then $e_{1}^{2}+e_{2}^{2}=$
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of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then $e_{1}^{2}+e_{2}^{2}=$
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Verified Answer
The correct answer is:
2
$e_{1}=\sqrt{1-\frac{b^{2}}{a^{2}}} \quad e_{2}=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$e_{1}^{2}+e_{2}^{2}=2$
$e_{1}^{2}+e_{2}^{2}=2$
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