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

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

Answer 2

The correct option is (B): 1

\(b^2=a^2(e^2−1)\)

\(⇒e^2−1=\frac{b^2}{a^2}\)

\(a^2=b^2(e^{'2}−1)\)

\(⇒e^{'2}−1=\frac{a^2}{b^2}=\frac{1}{e^2−1}\)

\(⇒(e^2−1)(e^{‘2}−1)=1\)

\(⇒e^2 e^{‘2}=e^2+e^{‘2}\)

\(⇒\frac{1}{e^2}+\frac{1}{e^{‘2}}=1\)