Question

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

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{5}{4}$

Answer 1

The correct option is A $\frac{3}{2}$

Eqn of ellipse

$\frac{x^2}{(\log a{)}^2}+\frac{y^2}{(\log b{)}^2}=1(a>b>0,a\ne 1)$

if $a>b>0$ then $\log a>\log b$

For ellipse

$e=\frac{1}{\sqrt{2}}$

$e=\sqrt{1-{\left(\frac{\log b}{\log a}\right)}^2}=\frac{1}{\sqrt{2}}\{\begin{array}{cc}as& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ e=\sqrt{1-\frac{b^2}{a^2}}& ifa>b\end{array}$

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

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

Eqn of hyperbola

$\frac{x^2}{({\log }_{b}a{)}^2}-y^2=1$

$e=\sqrt{1+\frac{b^2}{a^2}}\{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\}$

So

$e=\sqrt{1+\frac{1}{({\log }_{b}a{)}^2}}=\sqrt{1+\left(\frac{\log b}{\log a}\right)}\{as{\log }_{b}a=\frac{\log a}{\log b}$

$=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$e^2=\frac{3}{2}$