I like “Add a zero”, but lying in maths is wonderfully stupid. Read on !!!!!
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!
Diagram, then text:
The circle has radius 1 and centre at the origin.
The line is x = a
Now 1/a is the inverse of a, so a * 1/a = 1, and is fixed.
The line z from (1/a,0) to (x,y) is orthogonal to the radial line, so r/z = Y/a and the two triangles are similar
and r/(1/a) = a/R
Hence rR = a * (1/a) = 1, and both conclusions are true.
The point (x,y) follows a circular path, and rR = 1
This is a reblog.
Do read it.
It’s quite short!
See if you can read this to the end without answering the phone, noticing a notification, etcetera:
An area model, or a dot array model (same thing really) is one way of illustrating the algebraic completion of a square.
I have used dots as they are easier to create.
The quadratic is viewed initially as the “standard form”, and then rebuilt dynamically line by line into the “square plus a bit over” form, as shown in the following sequence:
The odd valued coefficient of x in the original expression can appear as a row and a column of half-dots, or half squares in the area model form.
Multiplication, the theory – by Thales’ theorem
The diagram can be simplified by using an acute triangle.
Proof of Thales theorem :
If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
Given : In ∆ABC , DE || BC and intersects AB in D and AC in E.
Prove that : AD / DB = AE / EC
Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.
1) EF ┴ BA 1) Construction
2) EF is the height of ∆ADE and ∆DBE 2) Definition of perpendicular
3)Area(ADE) = (AD.EF)/2 3)Area = (Base .height)/2
4)Area(DBE) =(DB.EF)/2 4) Area = (Base .height)/2
5)(Area(ADE))/(Area(DBE)) = AD/DB 5) Divide (3) by (4)
6) (Area(ADE))/(Area(DEC)) = AE/EC 6) Divide (3) by Area(DEC)
7) ∆DBE ~∆DEC 7) Both the ∆s are on the same base and
between the same || lines.
8) Area(∆DBE)=area(∆DEC) 8) So the two triangles have equal areas
9) AD/DB =AE/EC 9) From (5) and (6) and (7)
Not only this but also AD/AB = DE/BC
I borrowed this from http://www.ask-math.com/basic-proportionality-theorem.html
Some adjustments, but the Thales theorem is well done. I liked it.
To accommodate positive and negative numbers we need two extended number lines, with their zeros at the same place
Then multiplication of two negative numbers will always give a positive result, following the same geometrical structure.
The start, where the multiplier begins at 1
Now the 1 connects with the multiplicand -3
The multiplier is now placed at -2
And the parallel line from -2 connects to the 6 on the target line
This is so geometrical, and there is no “funny business”. None of the “ought to be 6”. No stuff about the distributive law.
The only geometry needs the Pythagoras theorem, and this will be the next post.
A number line is generally a piece of straight line with a starting point, labeled 0, and equally spaced points labeled 1 and 2 and 3 and 4 and so on till the paper runs out.
The value of a number is the distance from the zero point to the numbered point, in units of the equal spacing.
It is really much easier to draw one of these !
Two parallel number lines, same scale.
Notice that the zero points do not have to be in the same vertical line.
To get the symbolic form 7 – 2 = 5 we start with 0 on the target line (now the upper line) and join it to the 2 on the subtrahend line. (arrow down) (needs a nicer word here)
Then from the 7 on the subtrahend line we produce the line from 7 parallel to the 0 to 2 line. Then “arrow up” to the target line
Magic ! The result is 5 on the target line.
I like the picture, but the subtraction words are a mess.
We need two number lines, but since multiplication is “proportional” they will now be crossing, and the common point is labeled 0.
Also, the labels are “target” and “multiplier” and each line has its own scale.
Bonus: Nomograms, with lines.
The first is a simple calculator, with A + B = Sum
The second one calculates parallel resistances
Once upon a time, when I was deep into trigonometry, I followed the trail of sine and cosine sums, with sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
Then two more:
sin(A) + sin(B) = 2sin((A + B)/2)cos((A – B)/2)
and 2sin(A)cos(B) = sin(A + B) + sin(A – B)
Being unable to remember these last formulae (there are 8 altogether) I learned the basic one and derived, over and over again, the others.
In all of this I was never able to derive the basic formula until de Moivre’s theorem appeared.
So I had another go with trig and the simple version, and this is the result:
Aim of the game : sin(2x) = 2sin(x)cos(x)
First diagram:This does not look promising !
But what about sensible labelling –
The vertical from F meets AD in J
and the line from F at right angles to AB meets AB in K.
So we have two pairs of congruent triangles, BH and HD are both equal to sin(x),
and BHD is A STRAIGHT LINE
Now the vertical from B is sin(2x)
so sin(2x)/(2sin(x)) is the ratio which should be cos(x)