Math Is Fun Forum

JohnG

Novice

Registered: 2024-11-16

Posts: 3

need help finding intersections in Geogebra Classic

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

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,875

Re: need help finding intersections in Geogebra Classic

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!

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 10,618

Re: need help finding intersections in Geogebra Classic

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

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

JohnG

Novice

Registered: 2024-11-16

Posts: 3

Re: need help finding intersections in Geogebra Classic

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

Bob:  that was exactly what I expected to happen.

                           John

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,875

Re: need help finding intersections in Geogebra Classic

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)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

JohnG

Novice

Registered: 2024-11-16

Posts: 3

Re: need help finding intersections in Geogebra Classic

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

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,875

Re: need help finding intersections in Geogebra Classic

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.

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 (Today 14:47:10)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,875

Re: need help finding intersections in Geogebra Classic

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 (Today 23:23:10)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 10,618

Re: need help finding intersections in Geogebra Classic

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.

Consider y = ax^2 + bx + c

From the quadratic formula we can show that x = -b/2a goes through the lowest point (and is midway between the roots)

Let F be the point (-b/2a,  f) and let D be the line y = whoops I've got an algebraic error I'll finish this when I've sorted it out.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob