Example — Tangency to a parabola

For tangency to a curve one sets up a system of the line of the pencil with the curve and requires that the discriminant of the resolvent equation be zero. The generator excluded from the parametrisation must always be checked separately.

Example — Tangency to a parabola

In the pencil $f_k:\ y - 4 = m(x-0)$ with base point $P(0;4)$, determine the lines tangent to the parabola $y = x^2 - 2x - 2$.

Line–parabola system: $\begin{cases} y = mx + 4 \\ y = x^2 - 2x - 2 \end{cases} \implies x^2 - (2+m)x - 6 = 0.$

Tangency condition: $\Delta = 0$. $\Delta = (2+m)^2 + 24 = m^2 + 4m + 28.$ $\Delta = 0$ requires $m^2 + 4m + 28 = 0$ with $\Delta_m = 16 - 112 = -96 < 0$: no real value. The parabola is “too far” from the point $P$ — no tangent passes through $P$.

Excluded generator. For correctness, we check whether the vertical line $x=0$ (not present in the pencil parametrised by $m$) is tangent. Substituting $x=0$ into the parabola gives $y = -2$, a single point: $x=0$ is an improper secant, not a tangent. Hence the parabola has no tangents passing through $P$.

Solution: $\boxed{\text{no tangent line}}$.

Links

Topics: Pencils of lines
Concepts: Discriminant · Pencil of lines · Proper pencil · Pencil parameter · Tangency
Functions: Parabola · Line
Skills: Analytic geometry · Reasoning by cases · Solving systems