Question

If $e$ and $e^{{}^{\prime }}$ are the eccentricities of hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola, then the value of $\frac{1}{e^2}+\frac{1}{e^{{}^{\prime }2}}$ is

• A. 0
• B. 1
• C. 2
• D. none of these

Answer 1

Answer: (B)

If e is eccentricity $f$ hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $e=\sqrt{1+\frac{b^2}{a^2}}$.
Since, e is eccentricity of hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore e=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2}$
and e is eccentricity of hyperbola $\frac{x^2}{b^2}-\frac{y^2}{a^2}=1$
$\therefore e^{\prime }=\sqrt{1+\frac{a^2}{b^2}}$
$\Rightarrow (e^{\prime }{)}^2=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}$
$\therefore \frac{1}{e^2}+\frac{1}{(e{)}^2}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}$
$=\frac{a^2+b^2}{a^2+b^2}=1$