Question

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

• A. 0
• B. 1
• C. 2
• D. None of These

Answer 1

Answer: (B)

1

Answer 2

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 $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2}}=1$
$\therefore e'=\sqrt{1+\frac{a^{2}}{b^{2}}}$
$\Rightarrow(e')^{2}=1+\frac{a^{2}}{b^{2}}=\frac{a^{2}+b^{2}}{b^{2}}$
$\therefore \frac{1}{e^{2}}+\frac{1}{(e)^{2}}=\frac{a^{2}}{a^{2}+b^{2}}+\frac{b^{2}}{a^{2}+b^{2}}$
$=\frac{a^{2}+b^{2}}{a^{2}+b^{2}}=1$
Therefore, The Correct Option is (B): 1