Explain Intersection of Two Parabolas

By Robert O

After looking at the intersection between a parabola and a line in our previous tutorial, let’s now look at how we can find a solution to a system of two parabolas. Such a system has a solution if and only if they meet at one or more points. One solution suggests that the two parabolas are tangential to one another.

What does a parabolic equation look like?

The general equation of a parabola with vertex at the origin is 

    \[y=\frac{1}{4 p} x^{2}\]

. If the vertex is not at the origin, then the equation becomes 

    \[(y-k)^{2}=4 a(x-h)\]

where (h, k) is the vertex. We will explain different situations using examples.

Example 1

What are the intersection points of these two parabolas?


    \[y=-x^{2}+6 x-6.5\]


Equating right sides of the equations:

    \[-x^{2}+6 x-6.5=x^{2}-2\]

    \[2 x^{2}-6 x+4.5=0\]

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{6 \pm \sqrt{36-36}}{4}=1.5\]

Substituting x = 1.5 in the first equation:


The two parabolas intersect at only one point, which is (1.5, 0.25). That suggests that they are tangential to each other. Let’s find out from the graph.

Note that one solution does not always prove that the parabolas are tangential. You will need to graph them to prove the point. In this case, the two are tangential as the graph below shows.


Graph generated from https://www.desmos.com/

Example 2:

Find all the intersection points of the following parabolas:

    \[y=x^{2}-2 x+2\]

    \[y=-x^{2}+8 x+1\]


    \[x^{2}-2 x+2=-x^{2}+8 x+1\]

    \[2 x^{2}-10 x+1=0\]

Solve to find the value of x:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{10 \pm \sqrt{100-8}}{4}=\frac{10 \pm 9.592}{4}\]

    \[x=4.898 \text { or } 0.102\]

Substituting the x values in any of the equations result in:

    \[y=16.194 \text { or } 1.806\]

The two parabolas meet at two points: (4.898, 16.194) or (0.102, 1.806).

Let’s confirm our values by plotting the curves on a graph.

Graph generated from https://www.desmos.com/


The intersection of two parabolas is just like the intersection of linear lines. The only difference is that intersecting parabolas can results in many solutions. The procedure for finding these solutions is to first express the equations into standard forms, leaving you with only y on one side. Equating the other sides and solving for x gives you the coordinates of points of intersection of the parabolas.

About the Author

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

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