Intersections of two parabolas
Learning Objectives
- Set up the intersection of two parabolas as a system and eliminate \(y\) by subtraction.
- Recognize when the resulting equation is quadratic (\(a_1 \neq a_2\)) or linear (\(a_1 = a_2\)).
- Solve the system, finding all common points.
- Explain why two distinct parabolas meet in at most two points.
- Connect the count of intersections to the discriminant of the eliminated equation.
The last configuration
We have met a line with a parabola. The natural finale of the course is two parabolas with each other:
\[\begin{cases} y = a_1 x^2 + b_1 x + c_1 \\ y = a_2 x^2 + b_2 x + c_2, \end{cases} \qquad a_1, a_2 \neq 0.\]
As before, a common point is one whose coordinates satisfy both equations, so we solve the system. And as before, the method is elimination — but this configuration hides a small surprise that the line–parabola case did not, and it is worth watching for.
Subtracting removes the squares — sometimes
Both equations are solved for \(y\), so at a common point the two right-hand sides are equal:
\[a_1 x^2 + b_1 x + c_1 = a_2 x^2 + b_2 x + c_2.\]
Move everything to one side:
\[(a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0.\]
Here is the surprise. The coefficient of \(x^2\) is \(a_1 - a_2\), and this can be zero — precisely when the two parabolas have the same leading coefficient, \(a_1 = a_2\). Two genuinely different situations follow, and the first step of any such problem is to tell them apart.
\(a_1 \neq a_2\): the equation is a true quadratic. It has two, one, or zero solutions, by its discriminant — just like the line–parabola case.
\(a_1 = a_2\): the \(x^2\) terms cancel and what remains is linear, \((b_1 - b_2)x + (c_1 - c_2) = 0\), with at most one solution.
Two distinct parabolas meet in at most two points
Two distinct parabolas \(y = a_1 x^2 + b_1 x + c_1\) and \(y = a_2 x^2 + b_2 x + c_2\) have at most two common points.
More precisely, the abscissas of the common points are the real solutions of
\[(a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0,\]
and each common abscissa gives one common point. When \(a_1 \neq a_2\) there are at most two; when \(a_1 = a_2\) there is at most one.
A point \((x_0, y_0)\) lies on both parabolas iff \(y_0 = a_1 x_0^2 + b_1 x_0 + c_1\) and \(y_0 = a_2 x_0^2 + b_2 x_0 + c_2\). Subtracting the two equalities eliminates \(y_0\) and shows \(x_0\) must satisfy
\[(a_1 - a_2)x_0^2 + (b_1 - b_2)x_0 + (c_1 - c_2) = 0. \quad (\ast)\]
Conversely, any solution \(x_0\) of \((\ast)\) makes the two right-hand sides equal, so setting \(y_0\) to their common value gives a genuine common point. Common points therefore correspond exactly to real solutions of \((\ast)\).
It remains to see \((\ast)\) is not the trivial equation \(0 = 0\), which would have infinitely many solutions. If all three coefficients \(a_1 - a_2\), \(b_1 - b_2\), \(c_1 - c_2\) vanished, the two parabolas would have identical coefficients — the same parabola, contrary to being distinct. So at least one coefficient of \((\ast)\) is nonzero, making it a quadratic or a linear equation: at most two solutions when \(a_1 \neq a_2\), at most one when \(a_1 = a_2\). Either way, at most two common points.
The generic case: a quadratic remains
Find the common points of \(y = x^2\) and \(y = -x^2 + 2x + 4\).
Here \(a_1 = 1\) and \(a_2 = -1\) differ, so expect a quadratic. Setting the right-hand sides equal:
\[x^2 = -x^2 + 2x + 4 \iff 2x^2 - 2x - 4 = 0 \iff x^2 - x - 2 = 0.\]
This factors as \((x + 1)(x - 2) = 0\), so \(x = -1\) or \(x = 2\). Recovering \(y\) from the simpler parabola \(y = x^2\):
\[x = -1 \Rightarrow y = 1, \qquad x = 2 \Rightarrow y = 4.\]
The two parabolas meet at \((-1, 1)\) and \((2, 4)\). (Check in the other parabola: \(-(-1)^2 + 2(-1) + 4 = 1\) and \(-(2)^2 + 2(2) + 4 = 4\) — both agree.)
The surprise: a linear equation remains
Find the common points of \(y = x^2\) and \(y = (x - 2)^2\).
Both have leading coefficient \(1\): \(a_1 = a_2\). Watch the \(x^2\) terms cancel:
\[x^2 = (x - 2)^2 = x^2 - 4x + 4 \iff 0 = -4x + 4 \iff x = 1.\]
A single solution — the two parabolas meet in exactly one point, at \(x = 1\), \(y = 1^2 = 1\): the point \((1, 1)\). Note this is not a tangency: the parabolas genuinely cross there. Same leading coefficient means same "width and direction", so one parabola is just the other slid across — two such curves can cross only once.
Had we instead taken \(y = x^2\) and \(y = x^2 + 3\) (again \(a_1 = a_2\)), elimination would give \(0 = 3\), a false statement: no common point. Same shape, lifted straight up, never meeting.
Connection to Computer Science
Software that fits curves to data — a spreadsheet trendline, a scientific plotting tool — often models two different datasets as two parabolas and then asks where they cross: the break-even point between two quadratic cost models, or the moment two accelerating quantities become equal. The program finds it by subtracting the two fitted equations and solving what remains, checking first whether the leading terms cancel — the very case split of this lesson. The intersection it reports is one equation solved, exactly as here.
Why "at most two", and never more?
A line and a parabola met in at most two points; two parabolas, also at most two. It is tempting to expect that two curved shapes, each bending, might weave in and out and cross many times. Why does the count stay capped at two?
The reason is structural, and worth naming. Subtracting the two equations destroys a whole degree of the problem: the \(x^2\) terms, being the top of each parabola, either survive as a single \(x^2\) (one quadratic) or annihilate each other entirely (a line). The result can never be a cubic or higher, because there was no \(x^3\) anywhere to begin with — parabolas are degree two, and subtraction cannot manufacture a higher degree. A degree-two equation has at most two roots (Unit 1); a degree-one equation, at most one. The ceiling of two is inherited directly from the degree of the curves.
This is a first glimpse of a deep pattern: two algebraic curves of degrees \(m\) and \(n\) meet in at most \(m \cdot n\) points, and the eliminated equation's degree is what enforces it. For two parabolas the naive product would be \(2 \times 2 = 4\), but because both share the same variable \(y\) solved explicitly, the elimination collapses it to degree two. Counting intersections is really counting the roots of an eliminated equation — the thread that has run through this entire unit.
Exercises
Exercise 1
Find the common points of \(y = x^2 - 2x\) and \(y = -x^2 + 4\), listing abscissas smaller first.
Exercise 2
Eliminate \(y\) and count the real solutions.
Exercise 3
Decide whether each statement about two parabolas is true or false.
Exercise 4
Find the common point of \(y = x^2 + x\) and \(y = x^2 - 3x + 8\). (Notice the leading coefficients are equal.)
Exercise 5
For each pair of parabolas, decide whether eliminating \(y\) leaves a quadratic equation (leading coefficients differ) or a linear one (leading coefficients equal).
Summary
The common points of two parabolas solve their system; subtracting the equations eliminates \(y\) and leaves \((a_1 - a_2)x^2 + (b_1 - b_2)x + (c_1 - c_2) = 0\).
The case split: if \(a_1 \neq a_2\) the equation is quadratic (up to two common points, by its discriminant); if \(a_1 = a_2\) it is linear (at most one) — or impossible (none).
Two distinct parabolas meet in at most two points, a ceiling inherited from the degree of the eliminated equation; \(0 = 0\) would mean the parabolas coincide.
Solve by finding the abscissas, then recovering \(y\) from either equation.
The course, in one thread
From the first lesson to the last, one idea has done the work: bring a problem to a single second-degree equation and read its discriminant. Solving \(ax^2 + bx + c = 0\), locating a vertex, counting where a parabola crosses an axis, solving an inequality, meeting a line, meeting another parabola — each turned into a quadratic and was settled by \(\Delta\). The quadratic function is where algebra and geometry become two views of one object; you now hold both. The next courses build on it: the power function and radicals invert it, and trigonometry brings new functions to the same rigorous treatment.