Tag Archives: hyperbola

Analytic (coordinate) geometry has its good points, but elegance is not one of them.

This all started with a post by Maya Quinn (mathwater.wordpress.com) on problem solving. An oblong piece has been removed from an oblong cake (why? who did that?).The problem was to cut the remainder of the cake into two equal parts. Lots of solutions, one in particular was very imaginative.
This led me to another problem – what if the cut-out piece was triangular?

In view of one of the solutions to the first problem I decided that the triangle should be chopped in half. Not as simple as chopping a rectangle in half !
envelope half a cake

So, with Polya at my side (he’s been there since 1962) I decided that an equilateral triangle would be a reasonable starting point, as at least it was obvious that there were a few lines through the centroid doing the job (the medians), so some generality was still around.
Here is the equilateral triangle, nicely resting on the x axis, side length 2, top point on the y axis.
half equilateral 1

The vertical median bisects the triangle, so I described the general bisector GH by the distances x=DG and y=EH

In order to find out more about the bisector lines I first found a relationship between x and y, which was y=2x/(1+x), based on the area calculations:

Height of equilateral triangle is √3, base is 2, so area is √3

Height of BGH triangle is (2-y)/2 * √3, base is 1+x, so area is (1/2)*(1+x)*(2-y)/2*√3

and this is to be half of √3

So (1/2)*(1+x)*(2-y) must be equal to 1, and this leads to  y=2x/(1+x)
This allowed me to find the coordinates of the point H and locate it correctly on the line.

I then joined the points and by moving G the line moved, and I tracked it, shown in green below.

Small notational irritation: In the diagram below G is now C and H is now K
triangle area bisector detail
The visible curve is called the envelope of the lines, and it (obviously) touches the medians.

The complete envelope of the bisecting lines consists of two more sections, making a “concave”triangle with the centroid in the middle.

At this point I figured that a change of variable was in order, and looking at the y=2x/(1+x) equation, and at the diagram it looked like the distances of the two points from the left hand vertex would be helpful:

This produced X=1+x and Y=2-y, and to my surprise the resulting equation was Y=2/X, or XY=2.

The significance of this last equation escaped me at this point.

So I found the coordinates of the two points C and K in a coordinate system with origin at point A, in terms of the new variables X and Y, found the line joining them, rewrote in terms of a parameter P, used a bit of calculus to get the envelope (I’ll do a post on this later), and it had the second degree equation

√3/4 = y2 – √3y – √3xy + 3x

which is a hyperbola !!! (see C. Smith “Conic Sections”)

At this point I stopped thinking about halving a triangle, and looked at the full envelope for one of the three sections, and got this:
envelope hyperbola equilateral 2
A complete hyperbola, and, not only that, its asymptotes appear to be sides of the equilateral triangle. (Which when you think about it is not completely unreasonable !).

Then, thinking about doing shear operations on the picture, and with the X, Y variables, and ratios of lengths of segments on the same line being unchanged, I constructed the whole lot on an arbitrary triangle and did the envelope:
envelope hyperbola

The triangle is ADF and the halving line is BG. Yes ! Same result !

At which point I saw a bit of light, and thought that “XY=2” does not involve angles at all.

Ooops, there’s  another formula for the area of a triangle: a*b*sin(C)/2

Then straightaway all was revealed. You can draw the picture !
1. Triangle ACB, point X on side AC, point Y on side BC, area=0.5*AC*BC*sin(ACB)
2. Triangle XCY, area 0.5*XC*YC*sin(XCY), but angles ACB and XCY are the same, so if we require the area of triangle XCY to be half the area of triangle ACB we get 0.5*XC*YC*sin(XCY)=0.5*0.5*AC*BC*sin(ACB)

which reduces to (XC/AC)*(YC/BC)=0.5, and this does not involve the angle.

This corresponds to our earlier XY=2 since the equilateral triangle had side length 2

Pure speculation suggests that this result may have some connection with the way that the angle bisector of an angle in a triangle divides the opposite side in the ratio of the two adjacent sides.
———————————————–
The geometric diagrams were all constructed with the web based program
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html
√3

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Duality, fundamental and profound, but here’s a starter for you.

Duality, how things are connected in unexpected ways. The simplest case is that of the five regular Platonic solids, the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron. They all look rather different, BUT…..

take any one of them and find the mid point of each of the faces, join these points up, and you get one of the five regular Platonic solids. Do it to this new one and you get back to the original one. Calling the operation “Doit” we get

tetrahedron –Doit–> tetrahedron –Doit–> tetrahedron
cube –Doit–> octahedron –Doit–> cube
dodecahedron –Doit–> icosahedron –Doit–> dodecahedron

The sizes may change, but we are only interested in the shapes.

This is called a Duality relationship, in which the tetrahedron is the dual of itself, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other.

Now we will look at lines and points in the x-y plane.

3x – 2y = 4 and y = (3/2)x + 2 and 3x – 2y – 4 = 0 are different ways of describing the same line, but there are many more. We can multiply every coefficient, including the constant, by any number not 0 and the result describes the same line, for example 6x – 4y = 8, or 0.75x – 0.5y = 1, or -0.75x + 0.5y + 1 = 0

This means that a line can be described entirely by two numbers, the x and the y coefficients found when the line equation is written in the last of the forms given above. Generally this is ax + by + 1 = 0

Now any point in the plane needs two numbers to specify it, the x and the y coordinates, for example (2,3)

So if a line needs two numbers and a point needs two numbers then given two numbers p and q I can choose to use then to describe a point or a line. So the numbers p and q can be the point (p,q) or the line px + qy + 1 = 0

The word “dual” is used in this situation. The point (p,q) is the dual of the line px + qy + 1 = 0, and vice versa.

dual of a rotating line cleaned up1

The line joining the points C and D is dual to the point K, in red.  The line equation is 2x + y = 3, and we rewrite it in the “standard” form as  -0.67x – 0.33y +1 = 0  so we get  (-0.67, -0.33) for the coordinates of the dual point K.

A quick calculation (using the well known formula) shows that the distance of the line from the origin multiplied by the distance of the point from the origin is a constant (in this case 1).

The second picture shows the construction of the dual point.

dual of a rotating line construction1

What happens as we move the line about ? Parallel to itself, the dual point moves out and in.

More interesting is what happens when we rotate the line around a fixed point on the line:

gif duality rotating line

The line passes through the fixed point C.  The dual point traces out a straight line, shown in green.

This can be interpreted as “A point can be seen as a set of concurrent lines”, just as a line can be seen as a set of collinear points (we have fewer problems with the latter).

It gets more interesting when we consider a curve. There are two ways of looking at a curve, one as a (fairly nicely) organized set of points ( a locus), and the other as a set of (fairly nicely) arranged lines (an envelope).

A circle is a set of points equidistant from a central point, but it is also the envelope of a set of lines equidistant from a central point (the tangent lines).

So what happens when we look for the dual of a circle? We can either find the line dual to each point on the circle, or find the point dual to each tangent line to the circle. Here’s both:

dual of a circle4

In this case the circle being dualled is the one with center C, and the result is a hyperbola, shown in green.  The result can be deduced analytically, but it is a pain to do so.

dual of a circle3

The hyperbola again.  It doesn’t look quite perfect, probably due to rounding errors.

The question remains – If I do the dualling operation on the hyperbola, will I get back to the circle ?

Also, why a hyperbola and not an ellipse ? Looking at what is going on suggests that if the circle to be dualled has the origin inside then we will get an ellipse. This argument can be made more believable with a little care !

If you get this far and want more, try this very heavy article:

http://en.wikipedia.org/wiki/Duality_(mathematics)

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What they didn’t tell you at school (conic sections)

They tell you that conic sections are exactly what the words describe : Slice a double cone, the edge of the slice is a conic section, parabola, hyperbola, ellipse.

Then they tell you that y = x^2 is a parabola, or that all second degree equations in x and y are conic sections, or worst of all, they come up with the focus/directrix definition.

NOBODY shows you how to get the equation from the sliced cone !!!!!!!!!!!!

Well, here goes – (the math is after the picture, and it is so simple)

Image

Image

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Equations of conics

From the previous post (it is now the next one !) it can be seen that projection of the unit circle from the double cone vertex onto the slicing plane will give a second degree equation.  What a nice lead in to these equations, and also projective geometry, and to 3D xyz stuff.

The focus/directrix definition is so “rabbit out of a hat” that it’s time for its retirement. In any case, what use is the focus except for a parabola. And the directrix ? Surely “Directrix” is a female director !!!!!!!!!!!!

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