A line and a parabola
Learning Objectives
- Set up the intersection of a line \(y = mx + n\) and a parabola as a single quadratic equation.
- Solve the line–parabola system, finding the common points.
- Classify the relative position — secant, tangent, exterior — from the sign of the combined discriminant.
- Find a parameter (slope or intercept) that makes a line tangent, via the condition \(\Delta = 0\).
- Recognize that a vertical line always meets the parabola once, without being tangent.
Two curves in one plane
We have studied the line \(y = mx + n\) (in the linear course) and the parabola \(y = ax^2 + bx + c\) (in Unit 2). Draw both in the same plane and a new question appears: where — and whether — do they meet?
A point lies on both curves exactly when its coordinates satisfy both equations at once. So the meeting points are the solutions of the system
\[\begin{cases} y = ax^2 + bx + c \\ y = mx + n. \end{cases}\]
The pictures below show the three things that can happen — the line can cut the parabola in two points, touch it at one, or miss it entirely — and the whole story turns out to be decided, once again, by a discriminant.
One substitution collapses the system
Both equations give \(y\) in terms of \(x\), so the two right-hand sides must be equal at a common point:
\[ax^2 + bx + c = mx + n.\]
Bringing everything to one side gathers it into a single quadratic equation in \(x\) alone:
\[ax^2 + (b - m)x + (c - n) = 0.\]
Its leading coefficient is \(a \neq 0\), so this is a genuine quadratic equation — precisely the object of Unit 1. Each of its real solutions \(x_0\) gives one meeting point, whose ordinate is recovered from the line, \(y_0 = mx_0 + n\). Counting the meeting points is therefore counting the real roots of this equation: a job for its discriminant.
Secant, tangent, exterior
Let \(\Delta\) denote the discriminant of the combined equation \(ax^2 + (b - m)x + (c - n) = 0\). According to how a line meets a parabola, we say the line is:
secant if it meets the parabola in two distinct points (\(\Delta > 0\));
tangent if it meets the parabola in exactly one point (\(\Delta = 0\));
exterior if it does not meet the parabola at all (\(\Delta < 0\)).
The single point of a tangent line is called the point of tangency.
The discriminant decides the position
Let a line \(y = mx + n\) and a parabola \(y = ax^2 + bx + c\) (\(a \neq 0\)) be given, and let
\[\Delta = (b - m)^2 - 4a(c - n)\]
be the discriminant of their combined equation. Then the line is secant to the parabola if \(\Delta > 0\), tangent if \(\Delta = 0\), and exterior if \(\Delta < 0\) — meeting it in two, one, or zero points respectively.
A point \((x_0, y_0)\) lies on both curves iff \(y_0 = mx_0 + n\) and \(y_0 = ax_0^2 + bx_0 + c\), which holds iff \(x_0\) satisfies \(ax_0^2 + (b - m)x_0 + (c - n) = 0\) and \(y_0\) is then read off the line. So the meeting points correspond exactly — one to one — to the real roots of the combined equation.
That equation is quadratic, since its leading coefficient \(a\) is nonzero, and its discriminant is \(\Delta = (b - m)^2 - 4a(c - n)\). By the root-counting theorem of Unit 1, it has two distinct real roots when \(\Delta > 0\), one (double) root when \(\Delta = 0\), and none when \(\Delta < 0\). Transporting this count back through the one-to-one correspondence gives two, one, or zero meeting points — the three positions.
A secant line: solving the system
Find where the line \(y = x + 2\) meets the parabola \(y = x^2\).
Set the right-hand sides equal: \(x^2 = x + 2\), i.e. \(x^2 - x - 2 = 0\). Here \(\Delta = 1 + 8 = 9 > 0\), so the line is secant; the roots are \(x = \dfrac{1 \pm 3}{2}\), that is \(x = -1\) and \(x = 2\). Recovering \(y\) from the line \(y = x + 2\):
\[x = -1 \Rightarrow y = 1, \qquad x = 2 \Rightarrow y = 4.\]
The two meeting points are \((-1, 1)\) and \((2, 4)\). (Reading \(y\) from the parabola instead, \((-1)^2 = 1\) and \(2^2 = 4\), gives the same points — a useful check that no arithmetic slipped.)
A tangency condition: solving for a parameter
For which value of \(n\) is the line \(y = x + n\) tangent to the parabola \(y = x^2 - 3x + 4\)?
Tangency is not something to guess from a drawing; it is the exact algebraic condition \(\Delta = 0\). Form the combined equation:
\[x^2 - 3x + 4 = x + n \iff x^2 - 4x + (4 - n) = 0.\]
Its discriminant is \(\Delta = (-4)^2 - 4(4 - n) = 16 - 16 + 4n = 4n\). Tangency requires \(\Delta = 0\), hence \(n = 0\). The tangent line is \(y = x\), and the point of tangency comes from the double root \(x = -\dfrac{-4}{2} = 2\), giving \((2, 2)\).
This is the characteristic use of the discriminant as a condition rather than a calculation: an equation in the unknown parameter, solved by setting \(\Delta = 0\).
Connection to Computer Science
When a game or simulation launches a projectile — a thrown grenade, a basketball, a shell — its path is a parabola, and a flat ledge or sloping ramp it might land on is a line. To predict the landing, the engine solves parabola-equals-line, a single quadratic, and reads the discriminant: two solutions means the arc passes through the surface (entering and exiting), one means it just grazes, none means it sails clear. Aiming assistants that show where a throw will land are solving exactly this line–parabola intersection, many times per second as you adjust the aim.
Does "one common point" always mean tangent?
Our definition calls a line tangent when it meets the parabola in exactly one point. Reasonable — but it hides an assumption, and a single example breaks it.
Consider the parabola \(y = x^2\) and the vertical line \(x = 1\). They share exactly one point, \((1, 1)\) — yet the line plainly crosses the parabola there rather than grazing it. It is not tangent in any geometric sense. What went wrong? A vertical line is not of the form \(y = mx + n\); it is \(x = k\), and substituting gives no quadratic at all — just \(y = k^2\), a single point for the single input \(k\). A vertical line always meets the graph of a function exactly once, tangent or not, because a function has exactly one value at each input.
So our theorem carries a standing hypothesis: the line must be non-vertical, \(y = mx + n\). Within that family, one common point really does mean tangency (a double root). Outside it, "one point" is the ordinary behaviour of any vertical line and says nothing about touching. The lesson in rigor: a classification is only as trustworthy as the hypotheses quietly holding it up — name them, and test them at the edges.
Exercises
Exercise 1
For each pair, form the combined equation and use its discriminant to count the common points.
Exercise 2
The line \(y = 2x + 3\) is secant to the parabola \(y = x^2\). Find the two common points, listing the abscissas smaller first.
Exercise 3
Decide whether each statement about a line and a parabola is true or false.
Exercise 4
Use the condition \(\Delta = 0\) on the combined equation.
Exercise 5
For the parabola \(y = x^2\), decide from the combined discriminant whether each line meets it (secant or tangent, \(\Delta \geq 0\)) or misses it (exterior, \(\Delta < 0\)).
Summary
The common points of a line \(y = mx + n\) and a parabola \(y = ax^2 + bx + c\) are the solutions of their system; substituting collapses it to one quadratic, \(ax^2 + (b - m)x + (c - n) = 0\).
Its discriminant \(\Delta = (b - m)^2 - 4a(c - n)\) decides the relative position: \(\Delta > 0\) secant (two points), \(\Delta = 0\) tangent (one), \(\Delta < 0\) exterior (none).
To solve the system: find the roots \(x_0\), then recover \(y_0 = mx_0 + n\) from the line.
To impose tangency, set \(\Delta = 0\) and solve for the unknown parameter — the discriminant used as a condition, not a computation.
The "one point ⟹ tangent" rule needs a non-vertical line: a vertical line \(x = k\) always meets the parabola once, by crossing. Next, the finale: two parabolas meeting.