To get the
pluscurve of some curve C, you look for where its tangents intersect the axes. Say a tangent intersects the axes at (a, 0) and (0, b). Then (a, b) lies on the pluscurve of C.
Getting the
minuscurve is a little trickier. I don't have a solid definition for it. However the basic idea is that, for any point (a, b)
≠ (0, 0) on some curve C you draw a line, called a
minusline, between (a, 0) and (0, b). Do this for all points (a, b) on C. The union of the points on all minuslines will look like one or several contiguous regions in the plane. The border of the regions is the minuscurve of C.
For example, the blue region in the picture below has an interesting border in the first quadrant. That border is a minuscurve. By the end of this post, you will be able to describe that shape algebraically!
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| A set of minuslines making up a contiguous region. |
An interesting idea is that the pluscurve of the minuscurve of C might be C alone. Is this always the case? I don't know.
The pluscurve of ax + by + c = 0
The tangents of the line are simply the line itself. Assuming the line is non-horizontal and non-vertical it intersects the axes at (-c/a, 0) and (0, -c/b). Thus the pluscurve of the line is the point (-c/a, -c/b).
The plus- and minuscurves of xy = 1
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| xy = 1 in black and its pluscurve in dashed red. |
The pluscurve seems to be xy = 4 whereas the minuscurve seems to be xy = 1/4. I used the nice cross-platform tool
GeoGebra to visualise this. These are the GeoGebra files I created:
plus,
minus. What about the more general case xy = c? Is the pluscurve xy = 4c? 4c²?
I'll now derive the answer to that question. See the figure below. H is the hyperbola xy = c. It is parametrised by the point P = (t, c/t). For every P there is a corresponding P⁺ which lies on the pluscurve H⁺ of H.
We find P⁺ by observing where the tangent intersects the axes. To do that we need the equation of the tangent. Some linear algebra and calculus yields \(\frac{-1}{2t}x + \frac{-t}{2c}y + 1 = 0\). The intersection points are then readily obtained as (2t, 0) and (0, 2c/t). This means \(P^+ = (2 t, \frac{2c}{t})\) and that H⁺ obeys \(x y = 2 t \frac{2c}{t} = 4c\).
The plus- and minuscurves of the unit circle
Once again I used GeoGebra. It seems that the first quadrant part of the pluscurve is one branch of the hyperbola \((x - 1)(y - 1) = (\sqrt{2} - 1)^2\) and that the curve is symmetric about both the x and y axes. The minuscurve appears to have no points on it. However, the minuscurve of the first quadrant portion of the unit circle does exist. It looks almost exactly like the ellipse \(x^2 - x y + y^2 - 3x - 3y + 2 = 0\) near the origin:
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| Trying to approximate the border of the shaded area with an ellipse. |
The minuscurve of x + y = k (a parabola!)
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| Minuscurve of x + y = 1 |
Let's assume that k > 0 so that the set of minuslines of the curve looks like the picture above. For now, we are interested in the region \(0 \leq x \leq k\). The minuslines may be parametrized as lines passing through (t, 0) and (0, k - t). The equation of the lines is \(y = \frac{(k - t) - 0}{ 0 - t} (x - t) + 0 = \frac{t - k}{t} (x - t)\). For every x we want to find the t that gives the greatest y. Therefore we differentiate y with respect to t and look for \(\frac{dy}{dt} = 0\). (If we weren't so lazy we would assure ourselves that the second derivative is negative.)
$$y = \frac{tx - t^2 - kx + kt}{t} = x + k - t - \frac{kx}{t}\\
0 = \frac{dy}{dt} = \frac{kx}{t^2} - 1\\
t^2 = kx\\
t = \sqrt{kx}\\
y = x + k - \sqrt{kx} - \frac{kx}{\sqrt{kx}} = x + k - 2\sqrt{kx} = (\sqrt{k} - \sqrt{x})^2\\
\sqrt{y} = abs(\sqrt{k} - \sqrt{x}) = \sqrt{k} - \sqrt{x}\\
\sqrt{x} + \sqrt{y} = \sqrt{k}$$
This is actually a parabola! The major axis is y = x. We can see this if we square both sides and rotate it 45 degrees. We expect to get an equation of the form y = ax^2 + b. So, first
$$x + 2\sqrt{x}\sqrt{y} + y = k\\
2\sqrt{x}\sqrt{y} = k - x - y\\
4xy = (k - x - y)^2\\
x^2 - 2xy + y^2 - 2kx - 2ky + k^2 = 0.$$
Then, after rotation (see
Wikipedia or page 466 of "Calculus" 7th ed. by Robert A. Adams), we have
$$2x^2 - 2\sqrt{2}ky + k^2 = 0\\
y = \frac{2x^2 + k^2}{2\sqrt{2}k}$$
Some closing thoughts
In the special cases we've tried, we have found that branches of plus- and minus curves of conics are also branches of conics. Because conics are rather simple being second degree equations and because
simple things have meaning, plus- and minuscurves have meaning. Chances are we will find an application for them.
