Tangent with Given Slope

Tangent to y² = 4ax with slope m ≠ 0. One tangent exists for each slope (unlike a circle, which has two). Point of contact is (a/m², 2a/m).

For line $y = mx + c$ to be tangent to $y^2 = 4ax$, substitute into the parabola:

$$(mx + c)^2 = 4ax$$ $$m^2x^2 + (2mc - 4a)x + c^2 = 0$$

For a tangent, this must have exactly one solution, so discriminant $= 0$:

$$(2mc - 4a)^2 - 4m^2c^2 = 0$$ $$4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0$$ $$-16amc + 16a^2 = 0 \implies c = \frac{a}{m}$$

The tangent is:

$$y = mx + \frac{a}{m}$$

Point of contact: Substituting $c = a/m$ back into $m^2x^2 + (2mc-4a)x + c^2 = 0$:

The single root is $x = a/m^2$, giving $y = mx + a/m = a/m + a/m = 2a/m$.

Point of contact: $\left(\dfrac{a}{m^2},\, \dfrac{2a}{m}\right) = (at^2, 2at)$ with $t = 1/m$. Consistent with the parametric tangent $ty = x + at^2$ (slope $1/t = m$).

Contrast with circles: A circle has two tangents of each slope; a parabola has exactly one. This reflects the parabola's "one open end."