Question

If the eccentricities of the hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1and\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ be e and $e_1$, then $\frac{1}{e^2}+\frac{1}{e_1^2}=$

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

Answer: (A), (B)

2

$e=\sqrt{1+\frac{b^2}{a^2}}\Rightarrow e^2=\frac{a^2+b^2}{a^2}$
$e_1=\sqrt{1+\frac{a^2}{b^2}}\Rightarrow e_1^2=\frac{b^2+a^2}{b^2}\Rightarrow \frac{1}{e_1^2}+\frac{1}{e^2}=1$