Pair of Tangents from an External Point

Combined equation of tangents from (x₁, y₁) where S = x²/a² − y²/b² − 1, S₁ = x₁²/a² − y₁²/b² − 1, T = xx₁/a² − yy₁/b² − 1.

For $x^2/a^2 - y^2/b^2 = 1$, define:

$$S \equiv \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1, \quad S_1 \equiv \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1, \quad T \equiv \frac{xx_1}{a^2} - \frac{yy_1}{b^2} - 1$$

Combined equation of the two tangents from $(x_1, y_1)$ with $S_1 > 0$ (external point):

$$SS_1 = T^2$$

This is a second-degree equation representing the two tangent lines through $(x_1, y_1)$.

External vs internal: For the hyperbola, a point can be:

• External ($S_1 > 0$): two real tangents exist
• On the curve ($S_1 = 0$): one tangent (at the point itself)
• Between the asymptotes but outside the curve: complicated — requires checking which branch
• Between the branches ($S_1 < 0$): no real tangents

The geometry is richer than for the ellipse or circle because of the two branches and asymptotes.

Asymptotes from $SS_1 = T^2$: When $(x_1, y_1) \to (0,0)$ (the centre), $S_1 = -1$, $T = -1$, and $SS_1 = T^2$ gives $-S = 1$, i.e. $S = -1$... this needs care. The pair of asymptotes is obtained as $S = 0$ (combined equation), which is a separate formula.