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Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9 y^{2}=9,$ then the ratio $a^{2}: b^{2}$ equals
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8: 1
Given equation of ellipses is $x^{2}+9 y^{2}=9$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
Here, $a=3, b=1$
$c=\sqrt{(3)^{2}-(1)^{2}}=\sqrt{8}$
$\therefore$ Eccentricity of ellipse, $e=\frac{c}{a}$ $e=\frac{\sqrt{8}}{3}$
$\therefore$ Eccentricity of hyperbola $=\frac{3}{\sqrt{8}}$
$\Rightarrow$
$1+\frac{b^{2}}{a^{2}}=\frac{9}{8}$
$\Rightarrow$
$\frac{b^{2}}{a^{2}}=\frac{1}{8}$
$a^{2}: b^{2}=8: 1$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
Here, $a=3, b=1$
$c=\sqrt{(3)^{2}-(1)^{2}}=\sqrt{8}$
$\therefore$ Eccentricity of ellipse, $e=\frac{c}{a}$ $e=\frac{\sqrt{8}}{3}$
$\therefore$ Eccentricity of hyperbola $=\frac{3}{\sqrt{8}}$
$\Rightarrow$
$1+\frac{b^{2}}{a^{2}}=\frac{9}{8}$
$\Rightarrow$
$\frac{b^{2}}{a^{2}}=\frac{1}{8}$
$a^{2}: b^{2}=8: 1$
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