If `eandbare ` the eccentricities of a hyperbola and its conjugate, prove that `1/(e^2)+1/(bare ^2)=1.`

Explanation

Let, `(x^2)/(a^2)​−(y^2)/(b^2)​=1---------------(1)`
and
`(y^2)/(b^2)​−(x^2)/(a^2)​=1---------------------(2)`
are two hyperbola conjugate to each other.

Also let, e and `e_1​` are the eccentricities of (1) and (2) respectively.

Then, `e^2=1+(b^2)/(a^2)`​ and `e_1^2​=1+(a^2)/(b^2)`​

`1/[1+(b^2)/(a^2)]+1/[1+(a^2)/(b^2)]`

=`(a^2)/(a^2+b^2)​+(b^2)/(b^2+a^2)`
​

=1 (Proved)