Parabola, it’s scarily simple…

No distances, no circles, and you can easily derive an equation.

Just a right angled triangle.

First, the definition of a parabola from the focus and directrix.

Pick a line, the directrix, and a point (B) not on that line (the focus):

Find the line at right angles, passing through a point (C) on that line.

Now find the line from B to C, and the midpoint of BC, which will be D.

Find the line at right angles to BC from D, and the intersection of this line and the vertical line, E, is a point on the parabola.

As point C is moved the parabola is traced out.

The picture is completed with the line BE. Check it!

 

 

Halving a triangle, follow-up number one, ellipse

The previous post is “Analytic (coordinate) geometry has its good points, but elegance is not one of them”, in which a formula for any line cutting a triangle in half was found. The envelope of the cutting line for one section (of three) of the triangle was found to be always a hyperbola, which got me thinking “How do I get an ellipse?”. Clearly not by cutting a triangle in half, which involved taking two points A’ and B’on adjacent sides of the triangle, and making the product of their distances from the point of intersection of the sides equal to half the product of the lengths of the two sides.

So we cannot take two distances along two lines from the same point, lets try two distances from separate points on two lines, and keep the product of the distances constant. Magic:

The base line is AF. Line DL is set parallel to AF. B and G are the two points of interest, where AB and DG are the two distances, and the envelope of the line BG is an ellipse which touches AF and DL. The really interesting thing about this is that the lines do not have to be parallel, and that as the points A and F are placed nearer and nearer to the point of intersection the ellipse becomes more and more hyperbolic at the nearby end.

The next post will be a different follow-up to the triangle halving.

Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

 

Vertex of a parabola – language in math again

Here are some definitions of the vertex of a parabola.

One is complete garbage, one is correct  though put rather chattily.

The rest are not definitions, though very popular (this is just a selection). But they are true statements

Mathwarehouse: The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a
parabola.
Mathopenref: A parabola is the shape defined by a quadratic equation. The vertex is the peak in the curve as shown on
the right. The peak will be pointing either downwards or upwards depending on the sign of the x² term.
Virtualnerd: Each quadratic equation has either a maximum or minimum, but did you that this point has a special name?
In a quadratic equation, this point is called the vertex!
Mathwords: Vertex of a Parabola: The point at which a parabola makes its sharpest turn.
Purplemath: The point on this axis which is exactly midway between the focus and the directrix is the “vertex”; the vertex is the point where the parabola changes direction.
Wikibooks: One important point on the parabola itself is called the vertex, which is the point which has the smallest distance between both the focus and the directrix. Parabolas are symmetric, and their lines of symmetry pass through the vertex.
Hotmath: The vertex of a parabola is the point where the parabola crosses its axis of symmetry

Scoring is 10 points for finding the garbage definition and 5 points for the correctish definition !!!! Go for it!

When I studied parabolas, back in 1958 or so (!) the parabola had an apex. So I checked the meaning of vertex, and found that the word was frequently misused.

Here is a good account: https://en.wikipedia.org/wiki/Vertex_(curve)

Basically a vertex of a curve is a point where the curvature is a maximum or a minimum (in non math terms, most or least curved).

Here are two fourth degree polynomials, one has three vertices and the other has five. The maximum curvature points are indicated. The minimum curvature points are at the origin for the first curve, and at the points of inflexion for the second curve (curvature = zero)

A hyperbola has two vertices, one on each branch; they are the closest of any two points lying on opposite branches of the hyperbola, and they lie on the principal axis. On a parabola, the sole vertex lies on the axis of symmetry. On an ellipse, two of the four vertices lie on the major axis and two lie on the minor axis.

For a circle, which has constant curvature, every point is a vertex.

The center of curvature at a (nice) point on a curve is the center of the closest matching circle at that point. This circle will usually lie “outside” the curve on one side of the point, and “inside” the curve on the other side. Look carefully at the picture. It is called the osculating or kissing circle (from the Latin).

The center of curvature can be estimated by taking two point close to the point of interest, finding the tangents at these points, and then the lines at right angles to them and through the points. the center of curvature is roughly at the point of intersection of these two lines

The diagram below shows this estimate, for the blue parabola, at the vertex.

Finally (this has gone on further than expected!) I found this delightful gif.

Real problems with conic sections (ellipse, parabola) part two

So suppose we have a parabolic curve and we want to find out stuff about it.

Its equation … Oh, we have no axes.

Its focus … That would be nice, but it is a bit out of reach.

Its axis, in fact its axis of symmetry … Fold it in half? But how?

Try the method of part one, with the ellipse. (previous post)

This looks promising. I even get another axis, for my coordinate system, if I really want the equation.

Now, analysis of the standard equation for a parabola (see later) says that a line at 45 deg to the axis, as shown, cuts the parabola at a point four focal lengths from the axis. In the picture, marked on the “vertical”axis, this is the length DH

So I need a point one quarter of the way from D to H. Easy !

and then the circle center D, with radius DH/4 cuts the axis of the parabola at the focal point (the focus).

Even better, we get the directrix as well …

and now for the mathy bit (well, you do the algebra, I did the picture)

Yes, I know that this one points up and the previous one pointed to the right !

All diagrams were created with my geometrical construction program, GEOSTRUCT

You will find it here:

www.mathcomesalive.com/geostruct/geostructforbrowser1.html

Real problems with conic sections (ellipse, parabola)

So there is an oval hole in a metal casting. It’s supposed to be an elliptical hole. Is it ????? How can we find out ?????
A good start would be to find the line which would be the major axis if it was elliptical. This turns out to be an engineering problem, not a mathematical one (I cannot see a way!). If the oval curve has an axis of symmetry then the method below will find it:

Firstly, get a computer picture of the oval.
Take two circles, of different radii, and push them along until each one touches the oval in two places.

The line joining the two centers will be the axis of symmetry if there is one (this can be shown mathematically).

The equation of an ellipse uses the lengths of the major and minor axes. Do it !

The closeness to elliptic can be assessed in various ways. Think of one.

next…..finding the focus of a parabolic shape