Points of Intersection of a Parabola and a Line

The substitution method, worked through three cases: 2 points, 1 point (tangent), and no intersection

Overview

This tutorial explains how to find the exact points of intersection between a parabola and a line by solving their equations algebraically.

The method: isolate \( y \) in the linear equation, substitute it into the parabolic equation to obtain a quadratic in \( x \), solve that quadratic, then substitute each \( x \) back to get the corresponding \( y \).

Whether the quadratic has two, one, or no real solutions is decided by its discriminant \( \Delta = b^2 - 4ac \). All three cases are worked below.

How Many Intersections? The Discriminant Tells You

Discriminant Number of intersections Geometric meaning
\( \Delta > 0 \)2Line cuts the parabola at two distinct points
\( \Delta = 0 \)1Line is tangent to the parabola
\( \Delta < 0 \)0Line misses the parabola entirely

Case 1 — Two Points of Intersection (\( \Delta > 0 \))

Example 1 — Parabola and line meeting at two points

Find the points of intersection of the parabola and the line:

\[ y = 2x^2 + 4x - 3, \qquad 2y + x = 4 \]
View Solution

Step 1 — Solve the line for \( y \):

\[ 2y + x = 4 \;\Rightarrow\; y = -\frac{1}{2}x + 2 \]

Step 2 — Substitute into the parabola:

\[ -\frac{1}{2}x + 2 = 2x^2 + 4x - 3 \]

Step 3 — Collect into a standard quadratic:

\[ 2x^2 + 4x + \frac{1}{2}x - 3 - 2 = 0 \] \[ 2x^2 + \frac{9}{2}x - 5 = 0 \]

Step 4 — Clear the fraction and apply the quadratic formula:

Multiply by \( 2 \):

\[ 4x^2 + 9x - 10 = 0 \]

Discriminant:

\[ \Delta = 9^2 - 4(4)(-10) = 81 + 160 = 241 > 0 \]

So there are two distinct real solutions:

\[ x = \frac{-9 \pm \sqrt{241}}{8} \]

Step 5 — Get each \( y \) from the line \( y = -\tfrac{1}{2}x + 2 \):

For \( x = \dfrac{-9 - \sqrt{241}}{8} \):

\[ y = -\frac{1}{2}\cdot\frac{-9 - \sqrt{241}}{8} + 2 = \frac{9 + \sqrt{241}}{16} + \frac{32}{16} = \frac{41 + \sqrt{241}}{16} \]

For \( x = \dfrac{-9 + \sqrt{241}}{8} \):

\[ y = -\frac{1}{2}\cdot\frac{-9 + \sqrt{241}}{8} + 2 = \frac{9 - \sqrt{241}}{16} + \frac{32}{16} = \frac{41 - \sqrt{241}}{16} \]

Step 6 — Exact coordinates:

\[ \left( \frac{-9 - \sqrt{241}}{8},\; \frac{41 + \sqrt{241}}{16} \right) \quad \text{and} \quad \left( \frac{-9 + \sqrt{241}}{8},\; \frac{41 - \sqrt{241}}{16} \right) \]

Approximate coordinates (\( \sqrt{241} \approx 15.524 \)):

\[ (-3.07,\; 3.53) \quad \text{and} \quad (0.82,\; 1.59) \]

Check (both points on the line):

\( 2(3.53) + (-3.07) \approx 4.00 \) ✓    \( 2(1.59) + 0.82 \approx 4.00 \) ✓

Graph of the parabola y = 2x² + 4x − 3 and the line 2y + x = 4 with their two intersection points marked
Figure 1. Two points of intersection.

Case 2 — One Point of Intersection (\( \Delta = 0 \), Tangent Line)

Example 2 — A tangent line

Find the point of intersection of the parabola and the line:

\[ y = x^2 - 4x + 5, \qquad y = 1 \]
View Solution

Step 1 — The line is already solved for \( y \).

Step 2 — Substitute into the parabola:

\[ 1 = x^2 - 4x + 5 \] \[ x^2 - 4x + 4 = 0 \]

Step 3 — Compute the discriminant:

\[ \Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0 \]

A zero discriminant means one distinct real solution:

\[ (x - 2)^2 = 0 \;\Rightarrow\; x = 2 \]

Step 4 — The corresponding \( y \)-value is already known from the line: \( y = 1 \).

Answer: the single point \( (2, 1) \) — the line \( y = 1 \) is tangent to the parabola at its vertex.

Case 3 — No Point of Intersection (\( \Delta < 0 \))

Example 3 — A line that misses the parabola

Find the points of intersection of the parabola and the line:

\[ y = x^2 + 3, \qquad y = 1 \]
View Solution

Step 1 — Substitute into the parabola:

\[ 1 = x^2 + 3 \] \[ x^2 = -2 \]

Step 2 — Look at the discriminant of \( x^2 + 0x + 2 = 0 \):

\[ \Delta = 0^2 - 4(1)(2) = -8 < 0 \]

Conclusion: the quadratic has no real solutions, so the line and parabola do not meet. (Geometrically, the parabola has vertex at \( (0, 3) \) and opens upward, so no point on it can have \( y = 1 \).)

Answer: no points of intersection.

Practice Exercises

For each pair, decide whether the line and parabola intersect at 2 points, 1 point, or none — then verify by substitution.

Exercise 1

\( y = x^2 - 2x \) and \( y = x - 1 \).

View Answer

Substitute: \( x - 1 = x^2 - 2x \) → \( x^2 - 3x + 1 = 0 \). \( \Delta = 9 - 4 = 5 > 0 \), so two points.

\( x = \dfrac{3 \pm \sqrt{5}}{2} \), then \( y = x - 1 \).

Exercise 2

\( y = x^2 + 2x + 3 \) and \( y = 2x + 2 \).

View Answer

Substitute: \( 2x + 2 = x^2 + 2x + 3 \) → \( x^2 + 1 = 0 \). \( \Delta = -4 < 0 \), so the line does not intersect the parabola.

Frequently Asked Questions

How do you find the points of intersection of a parabola and a line?

Solve the line for \( y \), substitute into the parabola to get a quadratic in \( x \), solve that quadratic, then substitute each \( x \) back into the line to get the corresponding \( y \).

How many times can a line intersect a parabola?

0, 1, or 2 times, decided by the discriminant of the quadratic: \( \Delta > 0 \) gives two intersections, \( \Delta = 0 \) gives one (tangent), \( \Delta < 0 \) gives none.

What does it mean if the line is tangent to the parabola?

The line touches the parabola at exactly one point. Algebraically, the resulting quadratic has a double root (zero discriminant).

Can a horizontal line intersect a vertical parabola twice?

Yes. Any horizontal line above the vertex of an upward-opening parabola (below, for a downward-opening one) intersects it at two symmetric points.

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Last updated: October 17, 2026