It is well known that the simple quadratic equation $y = ax^2 + bx +c$ can be represented by a parabola. However, the relationship between the coefficients $a$, $b$, and $c$ and the shape of the parabola is not immediately obvious. At first, it seems reasonable to assume that it is possible to identify lines of suitable lengths on the graph of the parabola to represent $a$, $b$, and $c$, but this quickly turns out to be fairly difficult. The difficulty can be understood by applying dimensional analysis to the equation of the parabola.

In the graph, the abscissae ($x$-values) and ordinates ($y$-values) must have a dimension of **length**.
This is shown in the notation of dimensional analysis as $[L]$.
But the terms to be added together in the quadratic equation must be homogeneous in dimension.
That is, all the terms being added must have the *same* dimension:
$$\begin{align*} y &= a x^2 {}+ b x {}+ c \\ {[L]} & = {[L]} {}+ {[L]} {}+ {[L]} \end{align*}$$

So each of the terms $a x^2$, $bx$, and $c$ must have the same dimension of $[L]$, or length. This implies that:

- $a$ has dimension $[L]^{-1}$
- $b$ is dimensionless
- $c$ has dimension $[L]$,

which is why it is difficult to represent $a$ and $b$ on the graph. However, all is not lost.

The latus rectum of a parabola is the chord that is perpendicular to its axis. The length of the latus rectum is equal to the inverse (reciprocal) of $a$. This length clearly has dimension $[L]$, so its inverse has dimension $[L]^{-1}$, as required. Another characteristic length of the parabola is its focal length, the distance from the focus to the vertex. The focal length is a quarter of the length of the latus rectum, and so is equal to $1/4a$.

The value of $b$ is equal to the slope of the tangent to the parabola at the point where it crosses the $y$-axis. This can be derived by differentiating the equation of the parabola to get $$\frac{dy}{dx} = 2ax + b$$ where $dy/dx$ is the slope of the parabola, which is equal to the slope of the tangent. When $x = 0$, at the $y$-axis, then $dy/dx = b$. Since the slope is the ratio of two lengths, $b$ is dimensionless, as required.

The value of $c$ is just the vertical displacement of the parabola. When $x=0$ is substituted into the parabola equation, the result is $y=c$, so $c$ is the height at which the parabola crosses the $y$-axis.

We can put all the elements together in the following dynamic graph, implemented as a Java applet. It is also available as a downloadable application. (To execute the applet, it will be necessary to set up Java security, as described in security setup.)

In this graph, the coefficients $a$, $b$, and $c$ can be controlled by the sliders at the right-hand side. The focal length $f$ can also be specified in its own slider, but since the focal length and the coefficient $a$ are interlinked by $4af=1$, the sliders are also interlinked. The focus itself is shown by $\times$, and the latus rectum is the horizontal chord through the focus.

The value of $a$ is represented by the length $\ell$ of the latus rectum, which is equal to $1/a$ or $4f$.

The value of $b$ is the slope of the tangent where the parabola crosses the $y$-axis. The tangent is extended to the ordinate $x=1$, which it crosses at a height $b$ above the point at which the parabola intercepts the $y$-axis.

The value of $c$ is simply the intercept of the parabola on the $y$-axis. So the height at which the tangent crosses the $x=1$ ordinate is actually $b+c$. By symmetry, the tangent crosses the $x=-1$ ordinate at height $c-b$.

The meaning of the coloured rectangles is explained below.

The graph above also contains components that are helpful in visualising the roots of the equation: $$a x^2 + b x + c = 0$$ That is, the values of $x$ for which $y = 0$. The roots are the points at which the parabola crosses the $x$-axis.

The usual form in which the roots of this equation are expressed is: $$x = \frac{- b \pm \sqrt{ b^2 - 4 a c }}{2 a} $$ This is sometimes expressed as $$x = \frac{- b \pm \sqrt{ \Delta }}{2 a}$$ where $\Delta = b^2 - 4 a c $ is called the discriminant for the equation. The equation has real roots if the discriminant is not negative: $\Delta \ge 0$. (When $\Delta=0$, there is a single degenerate root.)Dimensional analysis again shows that $\Delta$ is actually dimensionless: $b$, and therefore $b^2$, are both dimensionless, and $4 a c$ has dimension $[L^{-1}][L]$, which is also dimensionless. So the numerators of the above fractions are dimensionless. The denominator $2 a$ has dimension $[L^{-1}]$, giving the correct dimension $[L]$ for $x$.

So once again it is difficult to visualise the abstract dimensionless discriminant $\Delta$ on the graph.

To provide a more concrete representation, re-express the roots as: $$\begin{align} x &= \frac{b}{2 a} \pm \sqrt{\frac{b^2}{(2 a)^2} - \frac{4 a c}{(2 a)^2}} \\ &= \frac{b}{2 a} \pm \sqrt{ \left( \frac{b}{2 a} \right)^2 - \frac{c}{a}} \end{align}$$ We then note that the length of the latus rectum is $\ell = 1/a$, so the roots can be expressed as: $$\begin{align} x = \frac{b}{2 a} \pm \sqrt{ \left( \frac{b}{2 a} \right)^2 - c \ell} \end{align}$$

Now, in the graph above, we note that the axis of the parabola is at $x = b/2a$, and draw the light grey square of side $b/2a$ on that axis. The yellow rectangle on the latus rectum is of height $c$ and width $\ell$. Thus, the area of the light grey square is $(b/2a)^2$, and the area of the yellow rectangle is $c \ell$.

The discriminant property for the existence of real roots now becomes: $$\left( \frac{b}{2 a} \right)^2 - c \ell \ge 0 $$

To express the discriminant property without reference to $a$, replace it with $1/\ell$, and the discriminant property becomes: $$\left( \frac{\ell b}{2} \right)^2 - c \ell \ge 0 $$

This property is true if the area of the light grey square is not less than the area of the yellow rectangle. We can draw a dark gray square whose area is the difference between the light grey square and the yellow rectangle. Its area is $(b/2a)^2 - c \ell$, so its side has length $\sqrt{(b/2a)^2 - c \ell}$. In the graph, we see that one of the roots of the equation is at one vertex of the dark grey square on the $x$-axis when another is on the axis of the parabola. The other root is equidistant from the axis on the other side, but is not shown to avoid clutter. When the discriminant is zero, the vertex of the parabola just touches the $x$-axis, producing the single degenerate root.

It is clear by examining the underlying algebra that the dark grey square (the root square) is equal to the light grey square minus the yellow rectangle, but so far I have been unable to develop a geometry-only proof of this.

When $c=0$, the equation reduces to: $$\begin{align}a x^2 + b x & = 0 \text{ or} \\ (a x + b) x & = 0 \end{align}$$ and the obvious roots are: $$\begin{align}x &= 0 \\ x &= - b / a \end{align}$$ which correspond to: $$\begin{align}x = \frac{b}{2 a} \pm \frac{b}{2 a} \end{align}$$ Geometrically, the yellow rectangle vanishes, and the light and dark grey squares are identical in size, with sides $b/2a$.

Copyright © Peter Havercan, 2015. The parabola and the simple quadratic equation by Peter Havercan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.