If the eccentricity of the ellipse x2log a2+y2log b2=1 a>b>0, a, b≠1 is 12and 'e' be the eccentricity of the hyperbola x2logba2-y2=1, then e2 is greater than (where log x=ln x)

Eccentricity of ellipsse x2log a2+y2log b2=1 a>b>0, a, b≠1 is 121-logb2loga2=12⇒log b2=log a21-12 ⇒logba2=2 Eccentricity of hyperbola e=1+1logba2 e2=1+logab2=1+12=32

If the eccentricity of the ellipse x2log a2+y2log b2=1 a>b>0, a, b≠1 is 12and 'e' be the eccentricity of the hyperbola x2logba2-y2=1, then e2 is greater than (where log x=ln x)

Eccentricity of ellipsse x2log a2+y2log b2=1 a>b>0, a, b≠1 is 121-logb2loga2=12⇒log b2=log a21-12 ⇒logba2=2 Eccentricity of hyperbola e=1+1logba2 e2=1+logab2=1+12=32

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If the eccentricity of the ellipse $\frac{x^{2}}{{(\log a)}^{2}} + \frac{y^{2}}{{(\log b)}^{2}} = 1 (a > b > 0, a, b \neq 1) is \frac{1}{\sqrt{2}}$ and 'e' be the eccentricity of the hyperbola $\frac{x^{2}}{{(\log_{b} a)}^{2}} - y^{2} = 1, then e^{2}$ is greater than (where $\log x = \ln x)$

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$\frac{3}{2}$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{5}{4}$

answer is B, C, D.

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Detailed Solution

Eccentricity of ellipsse $\frac{x^{2}}{{(\log a)}^{2}} + \frac{y^{2}}{{(\log b)}^{2}} = 1 (a > b > 0, a, b \neq 1) is \frac{1}{\sqrt{2}}$

$\sqrt{1 - \frac{{(\log b)}^{2}}{{(\log a)}^{2}}} = \frac{1}{\sqrt{2}}$

$\Rightarrow {(\log b)}^{2} = {(\log a)}^{2} [1 - \frac{1}{2}] \Rightarrow {(\log_{b} a)}^{2} = 2 E c c e n t r i c i t y o f h y p e r b o l a e = \sqrt{1 + \frac{1}{{(\log_{b} a)}^{2}}} e^{2} = 1 + {(\log_{a} b)}^{2} = 1 + \frac{1}{2} = \frac{3}{2}$