need help finding intersections in Geogebra Classic

JohnG · 2024-11-16 06:58:37

hi!
I am new to Geogebra.

I have created a parabola in Classic - - locus of points equidistant from a Focus and Directrix. Works fine.

I have a line that I would like to find the intersection of with the parabola. I select the line and parabola and . . . nothing happens.

What am I doing wrong?

phrontister · 2024-11-16 22:12:45

Hi JohnG, and welcome to the forum!

I know some Geogebra, but not parabolas.

Anyway, I drew a parabola in Geogebra, and then a line crossing it.

I found 2 ways by which to create a Point on the line/parabola intersection.

Both require first clicking on the Intersection tool, and then either:
1. Clicking directly on the intersection; or
2. While holding down the Ctrl key (on my PC), clicking on the line and the parabola (not necessarily in that order).

Hope that works!

Bob · 2024-11-16 23:54:36

hi JohnG

Welcome to the forum.

I just did this:

What did you expect to happen?

Phro. y = x^2 is a simple example of a parabola.

Bob

JohnG · 2024-11-17 00:36:27

Gentlemen: thanks so much! I will do some experimenting today.

Bob: that was exactly what I expected to happen.

John

phrontister · 2024-11-17 02:14:17

Hi JohnG;

Hope your experimenting is going well!

I did some of my own today, but I've hidden my results (a Geogebra drawing I did + explanatory notes). When you're ready to have a peek, just click on the hide box.

I hope my hidden info helps, but have a go at things yourself first!

Last edited by phrontister (2024-11-17 02:44:38)

JohnG · 2024-11-17 03:18:11

I have tried two approaches:

1) used the built in function in Geogebra for creating a parabola (specify point and line). Then made a random line and tried to intersect it. Worked fine.

2) built my own parabola using a Focus, Directrix, then a point on the directrix, and an isosceles triangle. The point on the directrix then drove a Locus which gave me a parabola. I might call this the "From Scratch" approach.

BUT when I try to intersect a random straight line with it, and am asked to select 2 objects, the parabola is simply not recognized and things fall flat.

So perhaps there is something lacking about a Locus that makes it impossible for the software to find the point of intersection.

thank you!

John

phrontister · 2024-11-17 23:01:50

Hi John;

Does your parabola show up in the Algebra window? If it does, then I don't understand why Geogebra doesn't recognise it.

JohnG wrote:

The point on the directrix then drove a Locus which gave me a parabola.

How did you 'drive' it?

I can think of 3 methods (no doubt there are more) of drawing a parabola in Geogebra:

Method 1. Activate Geogebra's 'Show trace' feature on the apex of the isosceles triangle, and then draw the parabola by moving the triangle's point A along the directrix. However, no parabola entry appears in the Algebra window, indicating that Geogebra doesn't recognise it for interaction. The Intersect tool doesn't work on the parabola, and I think the trace drawing is just a non-interactive image.

This is what I got with the 'Show trace' parabola feature:

Method 2. Create the parabola with Geogebra's Parabola tool.

Method 3. Create the parabola by entering this equation into the Input box: (x-1.5)² = 4(y-1). Equation created from Focus (1.5,2), Directrix y = 0.

For methods 2 & 3, a parabola entry appears in the Algebra window (indicating interaction capability), and I used the Intersect tool to create intersection points F & G on the parabola (see below):

Last edited by phrontister (2024-11-20 14:47:10)

phrontister · 2024-11-20 15:25:39

Bob wrote:

Phro. y = x^2 is a simple example of a parabola.

Thanks, Bob...that opened my eyes to a new field for me.

So I've explored that further, and progressed to the drawings in my post #7.

For my method #3, I had initially just used the 1.28x² - 3.84x - 5.12y = -8 equation that appeared in the Algebra window for method #2, in which I used the Parabola tool.

However, that equation was way more horrible than I expected, given that I'd used simple plotting figures: Focus (1.5,2) and Directrix y=0! I've got no idea how to construct that equation, neither does it make any sense to me.

A good online tutorial showed me how to find the much simpler-looking equation (x-1.5)² = 4(y-1), and gave me some understanding on how to make such things (if they're simple enough, that is).

Last edited by phrontister (2024-11-20 23:23:10)

Bob · 2024-11-20 21:08:34

The defining property of a parabola is as follows.

There is a point called the focus.
There is a line called the directrix.
The parabola is the locus of all points such that the distance to the directrix is equal to the distance to the focus

focus (1.5, 2) directrix y = 0 (ie the x axis)

Let a point on the parabola have coordinates (x,y)

Equating these:

I shall explore some properties of parabolas based on the defintion.

Left part of diagram.

Let the focus be F. There must be at least one point on the parabola that is nearest to the directrix. Let that point be R and let S be the point on the directrix so that FR = RS. I know that RS must be perpendicular to the directrix (distance implies shortest distance) but I have not assumed that FRS is a straight line

But FR = RS, so the triangle FRS is isosceles. Let T be the midpoint of FS. Then FT = TS so T is a point on the parabola and it is nearer than R. Unless R=T in which case FRS is a straight line.

So R is unique and FRS is perpendicular to the directrix.

Second part of diagram.

Let P be a point on the parabola and Q the point on the directrix so that FP = PQ.

Reflect P and Q in the line FRS to give points P' and Q'.

FP' = P'Q' so P' is also on the parabola.

So FRS is a line of symmetry for the parabola.

Bob

phrontister · 2024-11-21 00:19:02

Thanks, Bob!

I really like the first part, the section prior to your further exploration.

The logic is easy to understand, and, unlike the online approach that I used (that results in the same equation as yours), you've used Pythagoras, which I already know.

The online method (though simple enough), requires learning & memorising some additional techniques...which I can do without!

I'll explore the other part later, maybe tomorrow.

Last edited by phrontister (2024-11-21 09:06:43)

Bob · 2024-11-21 01:34:37

I'm working on a proof that all curves of the form y = ax^2 + bx + c are parabolas.

Progress so far:
y = x^2 fairly easy.
y = ax^2 not too much harder.
y = ax^2 + bx
y = ax^2 + bx + c follows easily by translation.

Bob

Bob · 2024-11-21 22:59:24

I've found a way.

step 1. consider a parabola with F = (0,a) and D is y = -a

If (x,y) is the general point on this parabola

Set a = 1/4 and we have y = x^2, so this well known curve is a parabola.

step 2. let a = 1/(4A) => y = Ax^2 is also a parabola.

step 3. Transform this by vector (0,d) => y = Ax^2 + d is a parabola.

step 4. consider y = ax^2 + bx + c

let the bracketed term be d/a

By the vector transform X = x + b/(2a) this becomes y = AX^2 + d and so is also a parabola.

Thus all quadratics of the form y = ax^2 + bx + c are parabolas.

Bob

phrontister · 2024-11-25 10:32:40

Hi Bob,

Bob (post #9) wrote:

Let T be the midpoint of FS. Then FT = TS so T is a point on the parabola

Let FR extend to U for straight-line FRU, with FR = RU. Then RU = RS, so U is a point on the directrix!

Last edited by phrontister (2024-11-25 13:23:07)

phrontister · 2024-11-25 10:37:03

Hi Bob,

Bob (post #9) wrote:

Left part of diagram...
...So R is unique and FRS is perpendicular to the directrix.
Second part of diagram...
...So FRS is a line of symmetry for the parabola.

Redrawing the diagram so that FRS is straight......places F, T and U in their correct positions.

Here's a slow-motion video of that happening: FRS straightening.

Last edited by phrontister (2024-11-25 18:38:26)

Math Is Fun Forum

#1 2024-11-16 06:58:37

need help finding intersections in Geogebra Classic

#2 2024-11-16 22:12:45

Re: need help finding intersections in Geogebra Classic

#3 2024-11-16 23:54:36

Re: need help finding intersections in Geogebra Classic

#4 2024-11-17 00:36:27

Re: need help finding intersections in Geogebra Classic

#5 2024-11-17 02:14:17

Re: need help finding intersections in Geogebra Classic

#6 2024-11-17 03:18:11

Re: need help finding intersections in Geogebra Classic

#7 2024-11-17 23:01:50

Re: need help finding intersections in Geogebra Classic

#8 2024-11-20 15:25:39

Re: need help finding intersections in Geogebra Classic

#9 2024-11-20 21:08:34

Re: need help finding intersections in Geogebra Classic

#10 2024-11-21 00:19:02

Re: need help finding intersections in Geogebra Classic

#11 2024-11-21 01:34:37

Re: need help finding intersections in Geogebra Classic

#12 2024-11-21 22:59:24

Re: need help finding intersections in Geogebra Classic

#13 2024-11-25 10:32:40

Re: need help finding intersections in Geogebra Classic

#14 2024-11-25 10:37:03

Re: need help finding intersections in Geogebra Classic

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