You might mention that any old cone contains *all* the conic sections, e.g., ellipses of every size and eccentricity. Thus, through any ellipse fixed in space there are infinitely many cones. It isn't well known, but their apices all lie on a hyperbola through the two foci. And the apices of all the cones through this hyperbola lie on the original ellipse! gosper.org/conethm.png (Somebody should animate this.) —rwg On 2018-11-08 10:35, Henry Baker wrote:
You should definitely show them the "Dandelion" (actually, Dandelin!) Theorem, which I call the *ice cream cone* theorem. It proves the sum-of-the-distances theorem about ellipses w/o requiring analytic geometry.
When my plane geometry teacher showed it to us, it completely blew my mind!
In particular, I couldn't get over the fact that the ellipse was symmetrical, even though the spheres were of different sizes!
Even if the kids haven't yet had 3D geometry, they can still follow this easy theorem.
https://en.wikipedia.org/wiki/Dandelin_spheres
At 05:43 AM 11/8/2018, Cris Moore wrote:
I'm going to talk to my daughter's 8th grade class about conic sections.
I want to focus on foci (ha), and how curves with beautiful geometric descriptions also have nice algebraic descriptions in Cartesian geometry.
But I found it surprisingly tricky to work out examples.
Consider an ellipse with foci at (-1,0) and (+1,0), and define the set of points where the sum of its distances from these two is 4.
Using Pythagoras' theorem produces an equation with a bunch of square roots.
Squaring both sides eventually turns this into 3x^2 + 4y^2 = 12 but this takes a bunch of steps of algebra, and mysterious cancellations of 4th-order terms.
Similarly, it takes a fair amount of work to get from the hyperbola with foci at (+2,+2) and (-2,-2), where the difference in distances is 4, to the simple equation xy = 2.
Am I doing something wrong?
Is there an easier way to get from foci and distances to these simple quadratic equations - without recourse to canonical forms, linear transformations, polar coordinates etc.?
Of course, I then want to talk about light waves bouncing from one focus to another…
I'm not sure how to justify this without a little calculus.
- Cris
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Ah, very nice - so this hyperbola through the two foci describes all the places in space from which the eliipse looks like a circle. It would be cool to see a move flying through the eliipse this way, with shading for depth… Cris
On Nov 9, 2018, at 8:21 AM, Bill Gosper <billgosper@gmail.com> wrote:
You might mention that any old cone contains *all* the conic sections, e.g., ellipses of every size and eccentricity. Thus, through any ellipse fixed in space there are infinitely many cones. It isn't well known, but their apices all lie on a hyperbola through the two foci. And the apices of all the cones through this hyperbola lie on the original ellipse! gosper.org/conethm.png (Somebody should animate this.) —rwg On 2018-11-08 10:35, Henry Baker wrote:
You should definitely show them the "Dandelion" (actually, Dandelin!) Theorem, which I call the *ice cream cone* theorem. It proves the sum-of-the-distances theorem about ellipses w/o requiring analytic geometry.
When my plane geometry teacher showed it to us, it completely blew my mind!
In particular, I couldn't get over the fact that the ellipse was symmetrical, even though the spheres were of different sizes!
Even if the kids haven't yet had 3D geometry, they can still follow this easy theorem.
https://en.wikipedia.org/wiki/Dandelin_spheres
At 05:43 AM 11/8/2018, Cris Moore wrote:
I'm going to talk to my daughter's 8th grade class about conic sections.
I want to focus on foci (ha), and how curves with beautiful geometric descriptions also have nice algebraic descriptions in Cartesian geometry.
But I found it surprisingly tricky to work out examples.
Consider an ellipse with foci at (-1,0) and (+1,0), and define the set of points where the sum of its distances from these two is 4.
Using Pythagoras' theorem produces an equation with a bunch of square roots.
Squaring both sides eventually turns this into 3x^2 + 4y^2 = 12 but this takes a bunch of steps of algebra, and mysterious cancellations of 4th-order terms.
Similarly, it takes a fair amount of work to get from the hyperbola with foci at (+2,+2) and (-2,-2), where the difference in distances is 4, to the simple equation xy = 2.
Am I doing something wrong?
Is there an easier way to get from foci and distances to these simple quadratic equations - without recourse to canonical forms, linear transformations, polar coordinates etc.?
Of course, I then want to talk about light waves bouncing from one focus to another…
I'm not sure how to justify this without a little calculus.
- Cris
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Cris Moore