Constructing a tangent line to a point on the curve $y = x^3$ using a compass and straight edge?

Question

There are number of ways of constructing a tangent line to the curve $y = x^2$ using a compass and straight edge. Does anyone know of a way of constructing a tangent line to the curve $y = x^3$ using a compass and straight edge?

Answer 1

We know that $\frac d{dx}x^3 = 3x^2$. Thus, when you have a point $(x,x^3)$ on the graph, it's easy to go 1 unit to the right and $3x^2$ units up — provided we have the $y$-axis and know how wide 1 unit is (and assuming we are on a cartesian grid with the same scaling of $x$ and $y$.

This means when we habe $y=x^3$ drawn, say, on some paper, we have to

1. Find the symmetry point $(x=y=0)$.

2. Find the line $x=0$ and $y=0$.

3. Find the point $x=y=1$.

No 3. is easy once we have 2. because we find $(1,1)$ by intersecting the graph with the angle bisector $y=x$ of the $x$- and $y$-axes. Likewise, if we have 1. and 3. then it's easy to find 2.

As we have only the graph any nothing else (at least that's what the question is implying as only the graph is mentioned and no coordinate lines or special points), all we can do is picking 2 points $P$ and $Q$ on the graph at random and to construct the 3rd points $R$ which lies on the line through $P$ and $Q$. This reminds of the group law on elliptic curves. However, I don't see a way to reach $(0,0)$ and $(1,1)$ in a finite number of steps. The coordinates are random, an all we know is that with almost certainty the quotient $x_P/x_Q$ is transcendental.

And btw. I don't see how it would for a parabola, except you have already given points with given known, rational quotient of the points' coordinates.

Answer 2

This a solution to my own question. I am the source. I have not been able to find a solution in the literature.The construction can be validated by anyone using a geometry program like geogebra.

Compass and Straightedge method for constructing a tangent to a Cubic $y = ax^3 +bx^2+cx + d$ at point $A$. The blue line $AL$ is the required tangent.

Note: the method does not work if point A is at the point of inflection. I did not include all the steps requiring a compass as this would clutter the animation.

NEW ...Inaddition to the first animation, I added two more animations. The second one shows a construction for finding the point of inflection and the vertical axis. The third shows a construction for finding the tangent line (in blue) through a point on the curve as well as the tangent line (in magenta) through the point of inflection when the point of inflection and the vertical axis have already been determined.

Answer 3

My understanding is that you are given the graph of $y = x^3$ plotted in a Cartesian coordinate system and an arbitrary point on the graph, and you are to construct a line tangent to the graph at that point without using any knowledge of the location of the origin of the coordinates or the orientation of the axes.

You have already given a valid construction, so this answer will focus on how we can prove it is valid.

The construction uses the following fact:

If $A$, $B$, and $C$ are three distinct points on the curve $y = x^3$ with $x$-coordinates $x_A$, $x_B$, and $x_C$ respectively, then $A$, $B$, and $C$ are collinear if and only if $x_A + x_B + x_C = 0.$

Proof of "only if": suppose $A$, $B$, and $C$ are collinear. Let $y = ax + b$ be the equation of the line on which the three points lie. Then the cubic polynomial $x^3 - ax - b$ has distinct roots at $x_A$, $x_B$, and $x_C$, and therefore $x^3 - ax - b = (x - x_A)(x - x_B)(x - x_C)$. Multiplying out the expression on the right and setting coefficients of the polynomials on both sides equal, we find that the coefficient of $x^2$ on the right is $-(x_A + x_B + x_C)$ whereas the coefficient on the left is $0$, so $x_A + x_B + x_C = 0.$

Proof of "if": Suppose $x_A + x_B + x_C = 0.$ Let $y = ax + b$ be the equation of the line on which the points $A$ and $B$ lie. Then the cubic polynomial $x^3 - ax - b$ has roots at $x_A$, $x_B$, and some value $x_0$, and therefore $x^3 - ax - b = (x - x_A)(x - x_B)(x - x_0)$ and $x_A + x_B + x_0 = 0.$ Therefore $x_0 = -x_A - x_B = x_C.$ This implies that the line $AB$ intersects the curve $y = x^3$ at a point $P$ with $x$-coordinate $x_C$. But since $y = x^3$ is the graph of a function of $x$, there is only one point on the curve with $x$-coordinate $x_C$, namely $C$. Hence the line $AB$ intersects the curve at $C$ and the three points are collinear.

We also use the following fact:

If $A$ and $B$ are distinct points on the curve $y=x^3$ with $x$-coordinates $x_A$ and $x_B$ respectively, then the line $AB$ is tangent to the curve at $B$ if and only if $x_A + 2x_B = 0.$

The proof is similar to the previous proof, except that in the "only if" direction we assume the polynomial $x^3 - ax - b$ has a double root at $x_B$, whereas in the "if" direction we have to prove $x_B$ is a double root.

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For ease of exposition, in the following I will write $x_P$ to denote the $x$ coordinate of point $P$, where the letter $P$ can be replaced by the name of any point.

Then the construction works as follows.

Let an arbitrary point $A$ on the curve $y = x^3$ be given, with the condition only that $A$ is not the point $(0,0).$

Find a line $ABC$ that intersects the curve in two other points $B$ and $C$. Therefore $x_A + x_B + x_C = 0$. Let $D$ be the midpoint of $B$ and $C$; then $x_D = \frac12(x_B + x_C) = -\frac12 x_A$.

Similarly, find a different line $AEF$ intersecting the curve at two other points $E$ and $F$. Therefore $x_A + x_E + x_F = 0$. Let $G$ be the midpoint of $E$ and $F$; then $x_G = \frac12(x_E + x_F) = -\frac12 x_A$.

Therefore $x_G = -\frac12 x_A = x_D$ and the line $DG$ is the line $x = -\frac12 x_A$. (Note that the $x$-coordinates of $D$ and $G$ are not only equal, but they are derived by essentially the same procedure.)

Construct the line $AH$ through $A$ parallel to $DG$. Therefore $x_H = x_A$. Take a point $I$ on the curve distinct from $A$. Reflect the point $I$ through $H$ to produce the point $J$. Therefore $x_I + x_J = 2x_H = 2x_A.$ Construct a line through $J$ parallel to $AH$, intersecting the curve at $K$. Then $x_K = x_J$ and $x_I + x_K = 2x_A.$ If the line $IK$ intersects the curve at a third point $L$, then $x_L = -2x_A.$ (If there is not a third intersection, you have chosen $I$ so that $IK$ is tangent to the curve; try a different point $I$.)

Now construct the line $AL$. Since $x_L + 2x_A = 0,$ this line is tangent to the curve $y = x^3$ at $A.$

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Note that I specified that $A$ is not the point $(0,0)$ in the proof of the construction. If $A$ is $(0,0)$ then all lines through $A$ that intersect the curve in two other points intersect the curve in points that are symmetric about $A$. Therefore if we attempt to follow the construction as described, we will find that $D = G = A$. Therefore the points $D$ and $G$ do not determine a line.

This is an easily resolved difficulty, because we can simply start the construction at some point other than $A$ in order to find a line parallel to the $y$ axis. The real difficulty is that the tangent at $A$ does not pass through any other point on the curve, so we cannot construct a distinct point $L$ on the curve that determines the tangent line.

If we suppose that we are able to detect that $A$ is the midpoint of $B$ and $C$, then we could construct a line through $A$ parallel to the $y$ axis, then a line through $A$ perpendicular to the first line. But I'm not aware that the standard toolbox of compass-and-straightedge constructions includes the ability to recognize that you have arrived exactly at a previously given point.

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The construction for the curve $y = x^3$ generalizes to a construction for $y = p(x)$ for any cubic polynomial $p(x)$, provided that $A$ is not at the inflection point of the curve, because the graph of $y = p(x)$ is simply a translation of a linear transformation of the graph of $y = x^3$, and the essential steps of the construction (the pairs of parallel lines and the proportional positions of the named points along the named lines) are all preserved by linear transformations and by translations.