you've been lied to about parabolas

[@BoringExtrovert]

I remember high school 10 years ago when we had to derive the equation of a parabola. We only had to do the special case where the directrix is either horizontal or vertical though

[@yamikira6512]

In my highschool they didn't even tell us about this definition of parabola. They told us: "Here you have y=ax^2+bx+c, you can calculate something called delta and find zeros." That's all, literally.

[@siquod]

In a university physics class, I got no points for a homework exercise to prove that a parabolic mirror works. I used the classical definition of a parabola to argue that all light rays travel the same distance to the focal point, but they expected me to use the analytical geometry stuff and do lots of unenlightening calculations.

[@ultrametric9317]

Instead of this brute force way, just do a dilation, translation and rotation, of the parabola y = x^2. The slope of the directrix is then immediately found from the rotation angle, the tangent of it.

[@MichaelGrantPhD]

Very cool. I think it would have been interesting to use a parametric formula for the line instead; e.g., (x,y) = (x0,y0) + t(dx,dy). That way, you could handle all slopes, including vertical, in one derivation.

[@vinesthemonkey]

and that translates nicely to vectors where the line is just defined by its normal!

[@Mathymagical]

Explain please? What you wrote is the parameterized line, not the parameterized parabola.

[@nozomusho]

I think you'll save a lot of algebra if you use the nice formula for distance from a point (x, y) to a line ax + by + c = 0 is d = |ax + by + c| / sqrt(a^2 + b^2). in fact, the literal definition of parabola translates nicely in this form: Any point on the parabola, (x, y), is equidistance from the focus (r, s), as to the directrix ax + by + c = 0: so sqrt((x - r)^2 + (y - s)^2) = |ax + by + c| / sqrt(a^2 + b^2). squaring both sides yields an expression fairly synonymous to the one in this video

[@vinesthemonkey]

even better: get rid of the coordinates and use vectors and a line's normal

[@ArpegiusWhooves]

To make the solution more general, the line should be ax + by + c = 0; this will only change the left side of the equation to (ax + by + c)^2 / (a^2 + b^2); not much worse i think.

[@ingiford175]

With the requirement Abs(a) + Abs(b) > 0

[@KadenHellewell]

There's a sign error there at the end. The constant term (which is got labeled "e") should be -(a^2+1)(r^2+s^2)+b^2

[@ulieggermann4346]

yes, thats right!😊

[@yanceyward3689]

I started up studying math again 3 years ago by basically going back to the beginning as I had done it 36 years earlier (I was research chemist who mostly didn't need to use any of the math I learned up through differential equations), but I made one change- I did the Schaum's Analytic Geometry text rather than skip it and go straight to trigonometry from college algebra. It made a world of difference to me- it made understanding everything that followed it so much more intuitive me, and I think it is a major missing piece of basic maths teaching today, at least for students who intend to study some advanced math in college and beyond.

[@joesoderstrom3110]

I appreciate the “change of variables” written with a delta

[@MCMCFan1]

Just do a coordinate transform where the directrix is the new x-axis. This yields a simple equation where the inverse transform yields the result instantly.

[@camrouxbg]

I graduated high school in 1996. One of the things we had to do in math was understand the Standard Form of conic sections. Ax^2 + By^2 +Cx + Dy + Exy + F = 0. We learned the conics as their locus definitions, and by understanding them as literally slicing through a double-napped cone. Fast-forward to taking geometry classes in 2016-2018 for my education degree and finding that none of the students were aware of these definitions. I mean, that stuff stressed me out at the time, but it was certainly beautiful. I also remember getting really stressed out because I couldn't understand what the directrix was. Like... given a parabola, how am I to know what the directrix is? It's not so bad in most cases, except when E is non-zero and the parabola is skew or rotated. Aah... good times.

[@vinesthemonkey]

I learned it too in middle school around 2011. But later I learned working with vectors and normals is so much easier conceptually than mindless algebra (that has exceptions like vertical lines)

[@2kreskimatmy]

i never thought about it this way

[@reeeeeplease1178]

We get the edge case (of a sqrt function) by going to the limit a -> infinity, leaving us with x^2 = (x-r)^2 + (y-s)^2 Rearranging yields y = s +- sqrt(2xr - r^2)

[@siquod]

The "distance to the directrix" part would have been much simpler and more general if you had treated the line equation as an implicit equation of the form =0.

[@gp-ht7ug]

This was the way to derive the equation of the parabola I learned at high school

[@roberttelarket4934]

So what have we been lied about?

[@ratandmonkey2982]

not sure. Maybe that parabolas could be different. But, they are all the same with a simple transformation of variables.

[@pietergeerkens6324]

I'm old enough to have had, for my "Advanced Functions" course in senior year of high school, 8 months of rotating, dilating, and translating the conic sections: parabola, ellipse, hyperbola ( plus the degenerate cases).

[@snatcc]

You’ve been lied about paraBALLSshahaha(hehehe

[@charlesglidden557]

Jesus said to his apostles ; Heaven is like 3X squared plus 6x minus 1.. The apostles looked at each other confused until Peter explaind : Don't worry that is just on of Jesus' parabolas.

[@vinesthemonkey]

coordinates obscure what's happening more generally. use vectors and the distance from a point to the line is just the perpendicular which is the normal!

[@gerryiles3925]

At 7:58, you should have put +x instead of +a... and you fixed it on the next board...

[@martincohen8991]

I think the polar form of the straight line would make it easier, since it is easy to get the distance to the line.

[@vinesthemonkey]

I think ditch the coordinates altogether and just use vector math...

[@stephenhamer8192]

Amusing to do this using vector methods: OK, we're going to place our plane in 3D space so we can use the vector product. We shall also place our origin on the directrix, which we shall suppose has direction *d*, a unit vector. Yes, yes, this is not fully general, but we could define our directrix as lying on the points *c* and *c* + *d*, and apply a preliminary translation followed by a dilation, i.e., *x* -> (*x* - *c*)/d, to achieve the desired data Let the focus, F, be at *f*, let *x* be a general point, P, on the parabola, and let N be the foot of the perpendicular from P onto the directrix. We observe that F is not on the directrix, so *d*, *f* are linearly indept and (*d*, 0) x (*f*, 0) = (0, 0, det (*d*, *f*)) =/= *0* In particular, det (*d*, *f*) =/= 0, so a matrix with columns *d*, *f* (or non-zero multiples thereof) will be non-singular. This will be useful to us later on We also have |(*d*, 0)| = |*d*| = d = 1, etc and (*d*, 0).(*f*, 0) = *d*.*f* Then PF^2 = | *x* - *f* |^2 = x^2 - 2.*f*.*x* + f^2 and PN^2 = |(*x*, 0) x (*d*, 0)|^2 = |(*x*, 0)|^2.|(*d*, 0)|^2 - [(*x*, 0).(*d*, 0)]^2 = x^2 - (*x*.*d*)^2 Equating and rearranging, we have: 2.*f*.*x* - f^2 = (*x*.*d*)^2, call this (*) Now change co-ords using the affine transformation: *x* -> trans [ trans *d*, trans 2.*f* ] *x* - trans [ 0, f^2] = trans [ X, Y ] NOTE: trans = the transpose operation, turning row vectors into column vectors and vice-versa, Then X = *d*.*x* and Y = 2.*f*.*x* - f^2, and (*) becomes: Y = X^2 Can this possibly be right?

[@stephenhamer8192]

google messed-up my notation! - anything with an * attached is a vector

[@bjornfeuerbacher5514]

4:40 to 10:30 You could get that result _much_ quicker by using the Hesse normal form for the line. But probably that formula isn't so well-known?

[@goodplacetostop2973]

16:44

[@Ahmed-Youcef1959]

I always like your comment ,it is the only one that i understand at 100% 😀😀

[@goodplacetostop2973]

@@Ahmed-Youcef1959 😂😂😂

[@ivanklimov7078]

also you can easily construct the type of parabola excluded from the video, ones with a vertical directrix, if you just remember your high school parabola construction. just swap x and y and boom, a sideways parabola

[@dimitardimitrakov2841]

I was like whaaaat when realising vertical lines cant be represented by any linear function. And I am in my 40s claiming to be well equipped for school level math

[@TheDannyAwesome]

What sort of shapes do you get if you use a different metric on R^2 for these geometric definitions of parabolas etc.? What about in the p-adic numbers instead of R?

[@Happy_Abe]

Why does the y^2 term get a coefficient and the x^2 doesn’t I understand the algebra that shows that, but how do we get parabolas like y=ax^2 if there’s no coefficient there

[@wesleydeng71]

The formula of distance between a point and a line is well known. No need to do it from the scratch.

[@nadonadia2521]

You do not have to do all this calculus the distance between a point (x1,y1) and the line Ax+By+C=0, is d= lAx1+By1+Cl/sqrt(A²+B²). In our case, y=ax+b ax-y+b=0 A=a b=-1, C= b the distance is d= lAx+By+Cl/sqrt(A²+B²), d=lax-y+bl/sqrt(a²+1)

[@__christopher__]

I thought the classical definition of a parabole was as intersection of a cone with a plane to which exactly one mantle line is parallel.

[@gregdeboer1]

It's definitely directrices. Like the plural of matrix

[@carultch]

How do you spell the possessive of Descartes?

[@CTJ2619]

you kind of did a little hand waving when saying that the line intersects the directrix at 90 degrees - why is that?

[@ZipplyZane]

The directrix is defined the shortest line between a point the line. And the shortest distance between a point and a line will have a right angle. Personally, I can just kinda visualize this. But ff you need proof, check out the Wikipedia article "Distance from a point to a line."

[@charleyhoward4594]

at 5:16 - u mult. 2 column vectors ?? confusing ....

[@shishkabob984]

It's a dot product, also known as a scalar product

[@richardlongman5602]

It's directrices. When you said directrixes I heard the ghost of my first Latin professor screaming "Nullo modo fieri potest: inno way whatsoever is it possible" and slamming his keys on the desk. Marvelous Pavlovian conditioning.

[@kkanden]

i guess the teaching method worked!

[@dlevi67]

I wonder what your Latin professor would say to the wonderful habit of declining the American-Latin singular of _alumni_ as _alum._ Which is a hydrated aluminium sulfate.

[@nightytime]

I hear "matrixes" and "matrice" all the time, lol

[@mikeholt2112]

In English, it’s matrixes and directrixes.

[@kendebusk2540]

@@nightytime Yes, and every day on business news you hear about the indexes going up or down. Indices? Too high falutin' !

[@theelk801]

focus rs? are you a car guy?

[@theimmux3034]

all hail y = x^2, the one true parabola

[@JarppaGuru]

..or they did not know just believe what other said

[@nnaammuuss]

so... what's the _lie?_ 🤔

[@carultch]

I think the "lie" is that a parabola is defined as the shape of the graph of y=x^2, and scaling/shifting transformations thereof. Not entirely a lie, but just a special case of a parabola with a horizontal directrix.

[@nnaammuuss]

is that how a parabola is defined in school, not as a section of a cone? oh okay, then. And... rotating, scaling, shifting allowed, y=x² is sufficient, no?

[@carultch]

@@nnaammuussIt is sufficient, it's just a limited scope of all that a parabola can be.

[@carultch]

​@@nnaammuussThat's why I put "lie" in quotes. Another example of a "lie" like this, is when you learn that the antiderivative of 1/x is ln(|x|) + C. Even in just the real numbers, there's more to the picture, because the +C can be different on both sides of the singularity. It's not that it's really a lie to not tell you this, it's just that it rarely governs an application of integration, so it's good enough to just keep it simple and let the +C be the same on both sides.

[@nelson6702]

Much as I enjoy these videos I find the "you've been lied to" thing clickbait really annoying.

[@26IME]

😦

[@physnoct]

click bait title