Cfrgtkky

As suggested by @Moti and @YvesDaoust in this post, a simple way to identify the roots (red dots) of a parabola (given focus and directrix, blue) by means of straightedge and compass is to draw the circle with center in the focus and radius the distance $overline{HI}$ between the $x$-axis and the directrix.

Now, consider the Quadratic Formula

$$
color{red}{x_{pm}}=frac{-bpmsqrt{b^2-4a cdot mathbf{c} }}{2a}.
$$

In the above image (a part the roots), it is easy to spot the term $mathbf{c}$, i.e. the intercept of the parabola with the $y$-axis.

My question is:

How to geometrically illustrate the other various algebraic terms of the quadratic formula by means of this construction, is such a way that the algebraic relation results immediately evident?

With geometrically, I mean some visual intuition, based on such plot (or something similar), in the spirit of a "proof without words".

Thanks for your help!

Here's a slight re-packaging of notions from my previous answer.

$$|OQ_{pm}| ;=; |BB_{-}| pm |MQ_{+}| ;=;-frac{b}{2a} pm sqrt{frac{b^2}{4a^2}-frac{c}{a}} ;=; frac{1}{2a} left(;-b pm sqrt{b^2-4ac};right)$$

• The figure represents the scenario in which $a>0$, $bleq 0$, $cgeq 0$ (and thus that $|OM|geq |MQ_{pm}|$). Adjustments to accommodate various sign changes should be clear.




• Figure labels and calculations incorporate the fact that the latus rectum has length $1/a$.




• That $overline{OM} cong overline{BB_{-}}$ is my previous answer's Property 2. That these segments' common signed length is $-b/(2a)$ follows, as before, from the equation of the represented parabola, by calculating the difference in $y$-coordinates for points with $x$-coordinates $pm 1/(4a)$.




• The relation between the highlighted areas follows immediately from my previous answer's Property 1. Stripping away the trappings of the specific problem, we can state the area property as a general principle that @Andrea should appreciate:

Property 1a. If squares are erected upon axis-perpendicular semi-chords of a parabola, then the difference in their areas is the area of the rectangle bounded by those chords and the extremities of the parabola's latus rectum.

By the way, here's a proof-without-words for my previous question's Property 2, using Property 1a above.

This solution isn't nearly as self-evident as I like my illustrations to be, but there are some interesting ideas here.

I'll preface this by noting, in something of an echo of @Rahul's comment, that geometricizing $y=ax^2+bx+c$ is a little tricky, in that $a$, $b$, $c$ are dimensionally distinct. In the approach described below, we take $x$ and $y$ (and thus also the roots of the quadratic equation) to be represented by ($1$-dimensional) lengths; necessarily, we see that $c$ must also be $1$-dimensional, $b$ must be $0$-dimensional (a ratio), and $a$ must be ... $(-1)$-dimensional!

Suppose that the graph of $y=ax^2+bx+c$ represents an upward-facing parabola with vertex $V= (h,-k)$; that is, we take $a$ positive and $b$ non-positive. Let $f$ be the vertex-to-focus distance, $f := |VF|$. Let the parabola cross the $y$-axis at $C$, of distance $cgeq0$ from the origin (although it's less problematic here to allow $c<0$), and let the parabola cross the $x$-axis at $R_{pm}$, at distances $hpm s$ from the origin.

Some auxiliary points: Let the $x$-axis and parabola axis meet at $M$ (the midpoint of $R_{+}$ and $R_{-}$). Let the horizontal line through $V$ meet the $x$-axis at $k$, and let $S$ be the projection of $R_{+}$ onto that line (so $|VK|=h$, $|VS|=s$, and $|OK|=|SR_{+}|=k$). Also, let the lines $x=pm f$ meet the parabola at $B_{pm}$, and let $B$ complete the right triangle with hypotenuse joining those points. Points $A$ and $G$ are on the $y$-axis and parabola axis such that $|KA|=|VG|=4f$.

Given the above, the below happens to be an illustration of the Quadratic Formula:

As I mentioned: not nearly as self-evident as I like. The illustration relies on two interesting properties of parabolas that derive from the reflection property; I'll prove them later.

Property 1. If $P$ is a point on the ("vertical") parabola, then its horizontal displacement from the vertex is the geometric mean of $4f$ and its vertical displacement from the vertex.

The illustration includes two instances of this property in the form of a classical right-triangle construction of the geometric mean.

$$begin{align}
triangle AVC: &quad frac{|KV|}{|KA|} = frac{|KC|}{|KV|} quadtoquad |KV|^2=|KA||KC|quadtoquad h^2=4f(c+k) tag{1} \[6pt]
triangle GSM: &quad frac{|VS|}{|VG|} = frac{|VM|}{|VS|} quadtoquad
|VS|^2=|VG||VM| quadtoquad s^2=4fk tag{2}
end{align}$$

From these, we may conclude $s^2 = h^2 - 4fc$, so that the $x$-coordinates of $R_{pm}$ ---that is, the roots of the quadratic polynomial--- have the form
$$h;pm;sqrt{h^2-4fc} tag{3}$$

(As an aside: Let the circumcircle $bigcirc R_{+} R_{-} C$ meet the $y$-axis again at, say, $D$. Then the power of a point theorems, applied to the origin with respect to this circle, imply $$|OR_{+}||OR_{-}| = (h+s)(h-s) = ccdot 4f = |OC||OD|$$
If we could show independently that $|OD| = 4f$, then we could reason conversely to get $(3)$ without the separate geometric means. I don't see an obvious way to make that association, however ... although little about this approach is obvious.)

Now, $(3)$ bears a bit of a resemblance to the Quadratic Formula. To get it closer, we invoke another property:

Property 2. If $P$, and distinct points $Q_{+}$ and $Q_{-}$, are on a ("vertical") parabola, such that the horizontal displacement from $P$ to each $Q$ is $f$, then the vertical displacement between the $Q$s is the distance from $P$ to the axis of the parabola.

In the figure above, $C$ plays the role of $P$, and $B_pm$ the roles of $Q_{pm}$. Since our parabola represents $y=ax^2+bx+c$, we have that $B_{pm}$ is at (signed) distance $af^2pm bf+c$ from the $x$-axis; thus, the vertical displacement between them is simply the difference of these expressions. By Property 2, we can write
$$h = left(;af^2-bf+c;right) - left(;af^2+bf+c;right) = -2bf tag{4}$$
(Recall that $b$ is non-negative here.) Therefore, $(3)$ becomes
$$-2bf;pm;sqrt{4b^2f^2-4cf} tag{5}$$
which we can write as
$$2fleft(;-b pm sqrt{b^2-frac{c}{f}};right) tag{6}$$
In light of the "known" observation that $a = dfrac{1}{4f}$ (there's that $(-1)$-dimensionality we needed!), we see
$$frac{1}{2a}left(;-bpmsqrt{b^2-4ac};right) tag{7}$$
so that we do, in fact, have the Quadratic Formula. $square$

I'm a little disappointed in the algebraic manipulations required in this demonstration. Perhaps a second pass at the argument, drawing from some more sophisticated geometric properties of parabolas, will streamline things.

Here are proofs of the Properties ...

Property 1.

Here, $overline{DW}$ is the directrix of the parabola, so that $triangle PFD$ is isosceles. The reflection property of parabolas implies that the tangent at $P$ bisects the angle at $P$; it therefore also bisects base $overline{FD}$ at a point $M$ that, by a simple similarity argument, also serves as the midpoint of $overline{BV}$. From similar subtriangles within $triangle PMD$, we have
$$frac{|BM|}{|BD|}=frac{|BP|}{|BM|} quadtoquad left(frac12 qright)^2=fp quadtoquad q^2 = 4fcdot p$$
giving the result. $square$

Property 2.

Again, $overline{DW}$ is the directrix. This time, we use the reflection property relative to $P$ to conclude that the tangent at $P$ is perpendicular to $overline{FD}$. It is "known" that chord $overline{Q_{+}Q_{-}}$ is parallel to that tangent. With a little angle chasing, we find that we may conclude $triangle Q_{+}QQ_{-}cong triangle FWD$, and the property follows. $square$

As the Wikipedia article Power of a point indicates, the tangent to the circle from the origin distance squared is the product of the two roots, but this is just $,c/a.$

The coefficients $a,b,c$ of the quadratic equation $ax^2+bx^2+c=0$ are not very geometric, so let's work with some slightly different variables that do have a geometric meaning:
begin{align}
alpha &= -frac b{2a}, & beta &= -frac cb, & gamma &= c.
end{align}
In reverse order, $C=(0,gamma)$ is the $y$-intercept of the parabola, $B=(beta,0)$ is the point where the tangent through $C$ meets the $x$-axis, and $A=(alpha,0)$ is the point on the $x$-axis with the same $x$-coordinate as the parabola's focus. The parabola is specified via $alpha,beta,gamma$, and we need to find the points $P$ and $Q$ where it crosses the $x$-axis.

^{Blue: given data, gray: constructed, green: equal quantities, red: desired roots}

Denote the focus by $F$ and the intersection of the directrix and the $y$-axis by $D$.

1. Construct the line $CF$ using the property of the parabola that the tangent $CB$ bisects $angle OCF$. Obtain $F$ as the intersection of $CF$ and the vertical through $A$.




2. Obtain $D$ using the fact that $C$ is equidistant from $F$ and $D$. The directrix is the horizontal through $D$, and is at distance $|OD|$ from the $x$-axis.




3. Obtain $P$ and $Q$ as the points on the $x$-axis at distance $|OD|$ from $F$.

$P$ and $Q$ are equidistant from $F$ and the directrix, and so lie on the parabola.

To derive the quadratic formula from this, we take an additional step, which may or may not be acceptable from a pure Euclidean-geometry perspective: We note that moving $C$ along the $y$-axis doesn't change the location of the roots, since it just scales the parabola vertically about the $x$-axis. Therefore, we may choose $C$ freely to simplify the construction.

In particular, let us take $C=(0,beta)$. Then $angle OCB=45^circ$, so the line $CF$ is horizontal, and $F=(alpha,beta)$. Now $|CD|=|CF|=alpha$, so $|OD|=alpha-beta$. The right triangle $triangle AFP$ has hypotenuse $|FP|=|OD|=alpha-beta$ and vertical side $|AF|=beta$, so the horizontal side is $|AP|=sqrt{(alpha-beta)^2-beta^2}=sqrt{alpha^2-2alphabeta}$; the same is true for $|AQ|$. Therefore,
begin{align}
{|OP|,|OQ|} &= |OA| pm |AP| \
&= alpha pm sqrt{alpha^2-2alphabeta}.
end{align}

Plug in the values of $alpha$ and $beta$ from above, and you obtain the quadratic formula.