Building Parabolas

Exploration Starting Point #1:

Building Parabolas from Linear & Non-Linear Functions including a closer look at creating ‘tilted’ parabolas and computing their axis of symmetry and vertex

During a recent roam around the internet for materials on using Geogebra, I came across a very interesting article entitled The Parabola as the Locus of the Product of Two Lines: Building Functions from Functions. This paper explores the construction of a parabola as a product of two lines and focuses on how to do this dynamically with Geogebra. I found the mathematical aspects of creating a quadratic function from two linear functions and the visual relationships of the graphs of the three functions very intriguing. The article only looked at a couple of examples to illustrate the relevant features of Geogebra. I also found a TI-Nspire activity (created by TI) called Products of Linear Functions which similarly guides students in creating a dynamic document to investigate the connections between the graphs of two linear functions and the graph of the quadratic function which is the product of the two linear functions. Although at the very end of the Geogebra article there is an invitation for readers to investigate how to produce horizontal parabolas, both the Geogebra article and the TI-Nspire activity limit their investigations to fairly simple examples where products of linear functions create parabolas that have a vertical axis of symmetry. Always on the lookout for suitable topics for Maths SL & HL Explorations, I wondered about trying to multiply two linear functions to construct a parabola that would have an axis of symmetry that was not vertical or horizontal. Such parabolas are something that most students would not be very familiar with (they’re not functions; fail the ‘vertical line test’) – and also it can lead into looking at parabolas as a particular type of conic section. Conic sections – not an especially advanced topic – is not in the syllabus for either Maths SL or Maths HL, so an Exploration connected to this should have a good chance of being at a commensurate level of mathematics for either course.

What follows is just some of my own “playing around” with creating parabolas from lines. I think there might be more than one starting point for a student Exploration here.

Start with a pair of ‘simple’ linear functions: y=x+1 and y = x 4 (Figure 1)

Find the product of the two linear expressions: y = ( x + 1 ) ( x 4 )= x 2 3 x 4 (Figure 2)

Figure 1

Figure 2

Not an unexpected result … that if you start with any two linear functions with zeros of x=1 and x=4 and form a quadratic function from their product then you’ll always get a parabola passing through ( 0 , 1 ) and ( 0,4 ) and thereby having the same axis of symmetry x= 3 2 .

For example, consider y = 4 x+4 and y= 1 2 x+2 that have ( 0 , 1 ) and ( 0,4 ) as their respective x-intercepts. Their product gives the following: y = ( 4 x + 4 ) ( 1 2 x + 2 ) = 2 x 2 + 6 x + 8 (Figure 3)

Figure 3

Figure 4

This result is best demonstrated with a dynamic (animated) graph (Figure 4 & video below).

For any value of the constant a the general line y=ax+a has an x-intercept of ( 0 , 1 ) and for any value of the constant b the general line y = b x 4 b has an x-intercept of ( 0,4 ) . By graphing their product and changing the values of a and b we can see how the resulting parabola changes but that its x-intercepts and axis of symmetry remain constant.

It appears that the graph of any quadratic function that is constructed from the product of two linear functions always has a vertical axis of symmetry. It made me wonder about trying to construct a parabola that has an axis of symmetry that is not vertical or horizontal.

Looking again at the first parabola, y= x 2 3x4 , it is possible to factorise with a different method rather than the ‘standard' way as y= x 2 3x4=( x+1 )( x4 ) .

Another way of factorising it is to re-write the quadratic expression as a difference of two squares:

y = x 2 3 x 4 = x 2 ( 3 x + 4 ) 2 = ( x + 3 x + 4 ) ( x 3 x + 4 )

Each of these factors can be expressed as a non-linear function: y=x+ 3x+4 and y=x 3x+4

The union of the graphs of these two functions is an oblique (‘tilted’/rotated) parabola having an axis of symmetry that is not vertical or horizontal (Figure 5) but with the same zeros, x=1 and x=4 , as the ‘original’ quadratic function. A single equation for this oblique parabola can be constructed by multiplying the relations y x 3 x + 4 = 0 and yx+ 3x+4 =0 .

Figure 5

( yx 3x+4 )( yx+ 3x+4 )=0 x 2 2xy+ y 2 3x4=0

Since x 2 2 x y + y 2 3 x 4 = 0 is not a function most graphic display calculators (GDC) would not be able to graph it. Some devices/software have the capacity to graph any conic section, such as the TI-Nspire (Figures 6 &7).

Figure 6

Figure 7

Equations in the form a x 2 + b x y + c y 2 + d x + e y + f = 0 represent one of several conic sections – parabola, hyperbola or ellipse (including circle) depending on the values of a, b and c (and three ‘degenerate’ cases – point, line and pair of intersecting lines). The discriminant of the general conic equation is b 2 4 a c , and if it has a value of zero then the graph of the equation will be a parabola.

The oblique parabola x 2 2 x y + y 2 3 x 4 = 0 appears to have the same shape as the original parabola y = x 2 3 x 4 which is not too surprising since x 2 2xy+ y 2 3x4=0 was derived from factors of x 2 3 x 4 . To check this, I physically cut out the graph of y = x 2 3 x 4 and placed it on top of the graph of x 2 2 x y + y 2 3 x 4 = 0 - and it matched perfectly (Figure 8). The graph of x 2 2 x y + y 2 3 x 4 = 0 must be a transformation of the graph of y = x 2 3 x 4 . It would appear to be a combination of a rotation and a translation.

Figure 8

Figure 9

Rather than trying to determine the precise details of the transformation, I thought it would be interesting to find a way to compute the axis of symmetry and vertex for the ‘new’ transformed parabola. Finding the axis of symmetry and vertex for a standard parabola – such as the ‘original’ parabola y = x 2 3 x 4 is not difficult. Most students have been taught methods and/or formulas for this. For the parabola y = x 2 3 x 4 , its axis of symmetry is the vertical line x= 3 2 and its vertex is the point ( 3 2 , 25 4 ) (Figure 9).

But how does one find the same for the oblique parabola x 2 2 x y + y 2 3 x 4 = 0 ? It’s possible that this might be more easily performed if the equations for the parabola were expressed in polar form rather than Cartesian form. I’m not sure – and I did not want to start over with a different parabola expressed with a ‘nice’ polar equation. Finding the point where the derivative of the parabola’s equation is zero is effective for vertical parabolas. Using implicit differentiation (not in Maths SL), it is possible to find a rule for the derivative of x 2 2 x y + y 2 3 x 4 = 0 , but the slope of the tangent at the vertex of this oblique parabola is not zero since it is not a minimum or maximum. By thinking about the relationship between any line parallel to the axis of symmetry and the respective parabola, it became clear that there is an algebraic approach by which the axis of symmetry can be computed – and that this can then be used to find the vertex.

A line will intersect a parabola at either two points, one point or no points. If a line intersects a parabola at exactly one point then the line must either be tangent to the parabola or parallel to the parabola’s axis of symmetry. Consider the ‘original’ parabola y = x 2 3 x 4 . Any vertical line will intersect the parabola at only one point. If a vertical line is rotated even by a very small angle it will eventually intersect the graph of y = x 2 3 x 4 at a second point.

A line passing through the origin cannot be tangent to the graph of x 2 2 x y + y 2 3 x 4 = 0 . Thus, a line passing through the origin which intersects the parabola at only one point must be parallel to the axis of symmetry. A line through the origin has the equation y = m x . Substituting this into the equation for the parabola will produce an equation whose solutions (in terms of m) will be the x-coordinates of any intersection points. The task is to find the value of m that causes this equation to have only one solution. Substituting for y: x 2 2x( mx )+ ( mx ) 2 3x4=0( m 2 2m+1 ) x 2 3x4=0
There will only be one solution if m 2 2 m + 1 = 0 which occurs when m=1 . Thus, the slope of the axis of symmetry is +1 which means that the original parabola was rotated clockwise 45 ( π 4 radians ) . Since the line tangent to the vertex is perpendicular to the axis of symmetry, its slope must be 1 . To find the vertex just need to determine the point ( x,y ) where the derivative of x 2 2xy+ y 2 3x4=0 is 1 .

Using implicit differentiation, dy dx = 2x2y3 2x2y . Solving 2x2y3 2x2y =1 gives y=x 3 4 . Substituting this result into the equation for the parabola and solving gives x= 55 48 (work done on CAS calculator – Figure 10).

Figure 10

Figure 11

Substituting this value of x into the parabola’s equation and solving for y gives two values for y (Figure 11) which is expected since it can be seen from the graph that the vertical line x= 55 48 will intersect the parabola at two points. The vertex is the lower of the two, thus y= 91 48 ; and, thus, the coordinates of the parabola’s vertex are ( 55 48 , 91 48 ) ; expressed as decimals ( 1.1458 3 ¯ ,1.8958 3 ¯ ) .

The axis of symmetry will then be the line with a slope of +1 containing ( 55 48 , 91 48 ) . The CAS calculator (Figure 12) computes this to be y=x 3 4 [ same as solution to dy dx = 2x2y3 2x2y =1 ] .

Figure 12

Figure 13

The graph above (Figure 13) shows oblique parabola x 2 2xy+ y 2 3x4=0 with its axis of symmetry and vertex indicated.

Here follows a condensed repetition of the same steps for another example.

1. Start with two lines and construct a parabola (qudratic function) from their product:

y=3x+1 and y=x+2 y=( 3x+1 )( x+2 )=3 x 2 5x+2

standard algebra/calculus methods give axis of symmetry x= 5 6 and vertex ( 5 6 , 49 12 )

Note that the zeros of the quadratic function are x=2 and x= 1 3 .

2. Factorise the quadratic by rewriting it as a difference of two squares:

y=( 3 x 2 +5x2 )=[ ( 3 x ) 2 ( 5x+2 ) 2 ]=( 3 x+ 5x+2 )( 3 x 5x+2 )

3. Write each of the factors as relations in terms of x and y and multiply them to produce a ‘new’ quadratic equation in the form a x 2 +bxy+c y 2 +dx+ey+f=0

( y= 3 x+ 5x+2 )( y= 3 x 5x+2 )=( y 3 x 5x+2 =0 )( y 3 x+ 5x+2 =0 )

3 x 2 +2 3 xy y 2 5x+2=0

4. Find slope of axis of symmetry of ‘new’ oblique parabola by finding slope of line through origin that has only one intersection point with the oblique parabola.

3 x 2 +2 3 x( mx ) ( mx ) 2 5x+2=0( m 2 +2 3 m3 ) x 2 5x+2=0

m 2 +2 3 m3=0m= 3

Note: Hence this ‘new’ parabola is a transformation of the ‘original’ parabola consisting of a clockwise rotation of 30 ( π 6 radians ) along with a translation.

5. Find vertex by determining at what point ( x,y ) the derivative of the quadratic (found by implicit differentiation) is equal to opposite reciprocal (perpendicular slopes) of slope of axis of symmetry. Computer algebra is used to assist in the computation.

dy dx = 6x2 3 y+5 2 3 x2y 6x2 3 y+5 2 3 x2y = 1 3 y= 3 ( 8x+5 ) 8

solving for x:

solving for y:

From the graph, the y-coordinate of the vertex must be the larger of the two solutions for y. Thus, vertex of the parabola is ( 53 320 , 253 3 320 ) , or to six significant figures ( 0.165625,1.36940 ) .

6. Find axis of symmetry by determining line with relevant slope containing the vertex. Computer algebra is used again to assist in computation.

axis of symmetry: y= 3 ( 8x+5 ) 8 = 3 x+ 5 3 8

The graph below shows oblique parabola 3 x 2 +2 3 xy y 2 5x+2=0 with its axis of symmetry and vertex indicated.

Further thoughts / questions / extensions

The biggest unanswered question is what are the precise details of the transformation performed on the ‘original’ vertical parabola to create the ‘new’ oblique parabola. It’s very interesting that in the two somewhat random examples investigated here the rotations are “nice” angles of 45 and 30 . Is that just coincidence? I doubt it.

It appears that multiplying the relations found by factorising the original quadratic as a difference of two squares (step 3) always creates a quadratic – in the form a x 2 +bxy+c y 2 +dx+ey+f=0 – that is a parabola rather than some other conic section. There must be some aspect of the original quadratic function that ensures that b 2 4ac=0 guaranteeing that the result is a parabola.

It is easy to find an equation for a vertical parabola with a given vertex, but what about finding an equation for a non-vertical parabola with a given vertex and axis of symmetry.

A critical insight for finding the slope of the axis of symmetry of an oblique parabola (step 4) is the fact that a line that intersects the parabola at exactly one point, and is not tangent to the parabola, must be parallel to parabola’s axis of symmetry. It would be useful to formally prove this.

Related Materials

TI-Nspire activity: Products of Linear Functions

Ask Dr Math - Math Forum: Rotating Parabolas

YouTube video: Functions – Rotating Conic Sections – Part 1 of 4 (18 min)

Desmos (browser-based graphing calculator): interactive Rotating Parabola

Use link below to download printable version (with some formatting changes) of article above.

Exploration Starting Pt - Building Parabolas     

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