If $e$ and $e'$ be the eccentricities of a hyperbola and its conjugate then $\large\frac {1}{e^2}+\frac{1}{e^{12}}$ is equal to

$\begin{array}{1 1}(A)\;0 \\(B)\;1 \\(C)\;2 \\(D)\;none\;of\;these \end{array}$

1 Answer

$\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 b^2=a^2(e^2-1) (or) e^2=1+\large\frac{a^2+b^2}{a^2} \qquad= \large\frac{a^2+b^2}{a^2} Conjugate hyperbola \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$

1 answer

1 answer

1 answer