Let the eccentricity of the hyperbola frac{x^ | vedclass

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(A) The given equation of the ellipse is $x^{2}+9y^{2}=9$, which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Here, $a^{2}=9$ and $b^{2}=1$,

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$8:1$

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(A) The given equation of the ellipse is $x^{2}+9y^{2}=9$, which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Here, $a^{2}=9$ and $b^{2}=1$, so $a=3$ and $b=1$. The eccentricity $e_{e}$ of the ellipse is given by $e_{e} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}$. The eccentricity $e_{h}$ of the hyperbola is the reciprocal of $e_{e}$, so $e_{h} = \frac{3}{\sqrt{8}}$. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the eccentricity is $e_{h} = \sqrt{1+\frac{b^{2}}{a^{2}}}$. Squaring both sides, we get $e_{h}^{2} = 1+\frac{b^{2}}{a^{2}} = \frac{9}{8}$. Thus, $\frac{b^{2}}{a^{2}} = \frac{9}{8}-1 = \frac{1}{8}$. Therefore, the ratio $a^{2}:b^{2} = 8:1$.

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9y^{2}=9$. Then the ratio $a^{2}:b^{2}$ equals:

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