Projective view of conics and quadrics | Differential Geometry 9 | NJ Wildberger

[@dreznik]

the explanation regarding the elliptic view of the parabola was brilliant

[@supurnasinha2133]

Great lecturing style ! All the concepts have been brought out lucidly through examples.

[@행복한나무-i9r]

Thanks for the basic of projective geometry... you saved my life :)

[@michaeltebele3305]

The intuitive reason why the projection of a parabola is an ellipse, is because the parallel lines converge to the vanishing point faster than the parabola extends away from it

[@iangrant9675]

I think there's a lot more in the ruled surfaces developed by conformal mappings of the complex plane. See my recent comment on kzbin.info/www/bejne/qZmonGupr5aihdk

[@iangrant9675]

That doesn't explain why the derivatives in the projection are zero at the vanishing point.

[@Razzildnb]

Great explanation of homogenous polynomials

[@yuanbo_chen]

The parabola seen as an ellipse part was mindblowing

[@Pilouface95]

Thank you sir, very useful 👍

[@blackbelttt]

Great!

[@sidharthghoshal]

@31:54 the projection of the hyperbola on the sphere is a bit misleading. In particular the sphere should be located ABOVE the origin of (X,Y,Z) to get that elliptical image. If the sphere is CENTERED at (0,0,0) then the hyperbola will only project onto the UPPER hemisphere and you get a rather degenerate looking object. EDIT: I just realized the professor corrects the mistake @32:43

[@sam111880]

If one wants to plot the homogeneous equation of this talk in mathematica one can uses this. To get more of a visualization on the surface connected to the y=x^2 in projective space. ContourPlot3D[x^2 == y* z , {x, -100, 100}, {y, -100, 100}, {z, -100, 100}] .

[@iangrant9675]

32:48 there is a potentially fruitful connection here with stability of numerical solutions under this duality. For an admittedly vague idea of what I mean, see livelogic.blogspot.com/2019/09/the-weirdest-math-video-youll-see-today.html

[@melissapereira6957]

what about the quadrics?

[@sam111880]

And for second example ContourPlot3D[x^2 - y^2== z^2, {x, -100, 100}, {y, -100, 100}, {z, -100, 100}]. And the last one is ContourPlot3D[x^3 + y^3==3*x*y*z, {x, -100, 100}, {y, -100, 100}, {z, -100, 100}].

[@BRoDINGoSoN123]

I wish there was a donate button - you've saved my degree x

[@Cid2065]

www.patreon.com/njwildberger ;)