Question

if e and e1 are eccentricities of a hyperbola and its conjugate, show that 1/e 2 + 1/e1 2 = 1

Answer 1

Let, x²/a²-y²/b²=1---------------(1) and
y²/b²-x²/a²=1---------------------(2)  are two hyperbola conjugate to each other.
Also let, e and e₁ are the eccentricities of (1) and (2) respectively.
Then, e²=1+b²/a² and e₁²=1+a²/b²
∴, 1/e²+1/e₁²
=1/(1+b²/a²)+1/(1+a²/b²)
=1/{(a²+b²)/a²}+1/{(b²+a²)/b²}
=a²/(a²+b²)+b²/(a²+b²)
=(a²+b²)/(a²+b²)
=1 (Proved)