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The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then, the eccentricity e is
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Verified Answer
The correct answer is:
$>\sqrt{2}$
Let the length of conjugate axis $=b$
and the length of transvers axis $=a$ Given that, $b>a$
Now,$b^{2}>a^{2}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$e^{2}=\left(\sqrt{1+\frac{b^{2}}{a^{2}}}\right)^{2}>2$
$e>\sqrt{2}$
and the length of transvers axis $=a$ Given that, $b>a$
Now,$b^{2}>a^{2}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$e^{2}=\left(\sqrt{1+\frac{b^{2}}{a^{2}}}\right)^{2}>2$
$e>\sqrt{2}$
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