Parametric Point on the Hyperbola

Every point on x²/a² − y²/b² = 1 can be written as (a secθ, b tanθ). θ ∈ [0°, 360°) with θ ≠ 90°, 270°. The right branch corresponds to θ ∈ (−90°, 90°) and the left branch to θ ∈ (90°, 270°).

Set $x = a\sec\theta$, $y = b\tan\theta$. Substituting:

$$\frac{a^2\sec^2\theta}{a^2} - \frac{b^2\tan^2\theta}{b^2} = \sec^2\theta - \tan^2\theta = 1 \checkmark$$

Branch identification: $\sec\theta > 0$ for $\theta \in (-90°, 90°)$ — right branch. $\sec\theta < 0$ for $\theta \in (90°, 270°)$ — left branch.

Alternative parametrisation using hyperbolic functions:

$$x = a\cosh t, \quad y = b\sinh t, \quad t \in \mathbb{R}$$

This covers only the right branch ($\cosh t > 0$ always). Both branches require $x = \pm a\cosh t$.

The $\sec\theta$/$\tan\theta$ form is more commonly used in Indian board and JEE problems.

Chord joining $\theta_1$ and $\theta_2$: The chord through $(a\sec\theta_1, b\tan\theta_1)$ and $(a\sec\theta_2, b\tan\theta_2)$:

$$\frac{x}{a}\cos\frac{\theta_1-\theta_2}{2} - \frac{y}{b}\sin\frac{\theta_1+\theta_2}{2} = \cos\frac{\theta_1+\theta_2}{2}$$