Pair of Tangents from an External Point

Combined equation of the two tangents from external point (x₁, y₁) to y² = 4ax, where S = y²−4ax, S₁ = y₁²−4ax₁, T = yy₁−2a(x+x₁).

For $y^2 = 4ax$, define:

$$S \equiv y^2 - 4ax, \quad S_1 \equiv y_1^2 - 4ax_1, \quad T \equiv yy_1 - 2a(x + x_1)$$

The combined equation of the two tangents from external point $P(x_1, y_1)$ is:

$$SS_1 = T^2$$

Derivation outline: Any point $Q(h, k)$ lies on a tangent from $P$ if the line $PQ$ is tangent to the parabola. Substituting the parametric form of line $PQ$ into $y^2 = 4ax$ and setting the discriminant to zero yields a relationship that, when cleaned up, takes the form $S(Q) \cdot S_1 = T(Q)^2$.

Expanded form:

$$(y^2 - 4ax)(y_1^2 - 4ax_1) = [yy_1 - 2a(x+x_1)]^2$$

This is a second-degree equation in $x$ and $y$, representing a pair of lines through $P$.

Angle between the tangents:

The angle $\theta$ between the two tangents from $P(x_1, y_1)$ satisfies:

$$\tan\theta = \frac{2\sqrt{a(x_1 - a)\cdot y_1^2}}{y_1^2 - 4a(x_1+a)}$$

The tangents are perpendicular when $y_1^2 = 4a(x_1 + a)$, which is the equation of the directrix — confirming that perpendicular tangents to a parabola meet on the directrix.