Line tangent to the parabola

A line is tangent to the parabola when it “touches” it at a single point: algebraically this happens when the discriminant of the line–parabola system equals zero ($\Delta=0$).

Example

Given the parabola $y=\frac12 x^2-2x+1$ and the line $y=mx-3$, for which values of $m$ is the line secant, tangent or external?

Vertex of the parabola: $x_V=-\dfrac{-2}{2\cdot\frac12}=2$, $\ y_V=\frac12\cdot4-4+1=-1$, so $V(2,-1)$.

System: $mx-3=\frac12 x^2-2x+1 \implies \frac12 x^2-(2+m)x+4=0$.

Discriminant: $\Delta=(2+m)^2-4\cdot\tfrac12\cdot4=m^2+4m+4-8=m^2+4m-4.$

• $\Delta=0$ (tangent): $m^2+4m-4=0 \implies m=-2\pm2\sqrt2$;
• $\Delta>0$ (secant): $m<-2-2\sqrt2$ or $m>-2+2\sqrt2$;
• $\Delta<0$ (external): $-2-2\sqrt2<m<-2+2\sqrt2$.

Example — Pencil of lines through a point

Given the parabola $y=-\frac12 x^2+x+1$, which passes through $P(0,1)$, and the line $y=mx+1$ (pencil of lines through $P$): $mx+1=-\tfrac12 x^2+x+1 \implies \tfrac12 x^2-x(1-m)=0 \implies x\left[\tfrac12 x-(1-m)\right]=0.$ The solutions are $x=0$ (always, because the line passes through $P$) and $x=2(1-m)$. The line is tangent at a point other than $P$ when $2(1-m)=0$, that is $m=1$.

Links

Topics: Parabola
Concepts: Discriminant · Pencil of lines · Straight line · Tangent line · Vertex
Functions: Parabola
Skills: Analytic geometry · Solving equations · Solving systems