Question
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,prove that $\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1$.
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$f(x)=\frac{6 x-7}{3}$
$\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right|=0$
$\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$