Conjugate Hyperbola

The conjugate of x²/a² − y²/b² = 1 is x²/a² − y²/b² = −1, equivalently y²/b² − x²/a² = 1. They share the same asymptotes. If e₁ and e₂ are eccentricities of a hyperbola and its conjugate: 1/e₁² + 1/e₂² = 1.

The conjugate of $H: x^2/a^2 - y^2/b^2 = 1$ is $H': x^2/a^2 - y^2/b^2 = -1$, equivalently $y^2/b^2 - x^2/a^2 = 1$.

Shared asymptotes: $H$ has asymptotes $x^2/a^2 - y^2/b^2 = 0$. So does $H'$. Two hyperbolas sharing the same asymptotes are called conjugate hyperbolas.

Eccentricities: $e_1$ for $H$ satisfies $e_1 = c/a = \sqrt{1+b^2/a^2}$. $e_2$ for $H'$ (which has transverse axis $2b$) satisfies $e_2 = c/b = \sqrt{1+a^2/b^2}$.

$$\frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1$$

Combined equation of conjugate pair: The two hyperbolas $H$ and $H'$ together with their shared asymptotes satisfy:

$$\left(\frac{x^2}{a^2} - \frac{y^2}{b^2} - 1\right)\left(\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1\right) = \left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right)^2 - 1$$

In JEE problems: When a chord of $H$ is found, it often bisects the conjugate $H'$ — a standard locus result.