The intersection of Line and Parabola Example

By Robert O

We are already aware of the solution of a system of two linear equations. To find a solution to a system of linear equations, we apply the Gaussian elimination, algebraic substitution, or graphical method. We covered that in our previous tutorial.

How do I solve a system of parabolic and linear equations?

What if we have one parabolic equation and the other a linear equation? We cannot use the Gaussian elimination in this case, but the other two methods apply. In this tutorial, we will use algebraic substitution and confirm our solution by looking at the graphs.

Line and a parabola can either intersect at two points, one point or never intersect at all. If you have two solutions, then the system intersects at two points. One solution shows that only one intersection point exists, and the line may be tangent to the parabola. A none solution problem means that the line and parabola do not meet.


The equation of a line is of the form:

    \[y=m x+c\]

A parabolic equation takes the form:

    \[y=\frac{1}{4 p} x^{2}\]


Example 1


    \[y=-2 x-1\]

    \[y=x^{2}-4 x\]


The two equations are all in standard form. So, we solve for x by equating the right sides to each other.

    \[-2 x-1=x^{2}-4 x\]

    \[x^{2}-2 x+1=0\]

This is now a quadratic equation problem. By applying the formula:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{2 \pm \sqrt{4-4}}{2}=1\]

We only have one solution for x. Find y from the linear equation.


The line and the curve intersect at one point only, which is (1,-3). Let’s confirm the solution by graphing them.

Graph generated from

We have only one solution, and we can see that the line is tangential to the parabola at (1,-3).

Example 2:


    \[2 y=-x+4\]

    \[y=2 x^{2}+4 x-3\]


Rewriting the equation in standard form:

    \[y=-\frac{1}{2} x+2\]

Equating the right side to each other:

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

    \[2 x^{2}+4.5 x-5=0\]

The result is a quadratic equation. By applying the formula:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-4.5 \pm \sqrt{20.25+40}}{4}=\frac{-4.5 \pm 7.762}{4}\]

    \[x=-3.066 \text { or } 0.816\]

Substituting these values in the linear equation:

    \[y=3.533 \text { or } 1.592\]

The parabola and the line meet at two points (-3, 3.533) and (0.866, 1.592). Confirming our solution using a graph proves that our solutions are, indeed, correct.


Graph generated from

Example 3


    \[y=2 x+1\]



We start by equation the right side of both equations.

    \[2 x+1=x^{2}-2\]

    \[x^{2}-2 x-3=0\]

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2}\]

    \[x=3 \text { or }-1\]

Substituting these values in the linear equation gives the values for y.

    \[y=7 \text { or }-1\]

The curve and the line intersect at two points: (3,7) and (-1,-1). Graphically, we get the same solutions.

Graph generated from


There is no challenge in finding solutions to systems of parabolas and lines. Just rewrite the equations in standard form and equate the right sides to each other. Also, note that you can have a parabola taking any shape. It can be opening upwards, downwards, left, or right. The line can also have either a negative or positive gradient (slope). Just solve the problem the way it is. Nothing changes.

About the Author

This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.