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Is there geometric proof of Cartesian equations (or rather geometric proof that implies Cartesian equation) of conic sections, ellipse, hyperbola and parabola which use their standard definition (ellipse is a curve for which $r_1 + r_2 = 2a$, simillary for hyperbola $|r_1 - r_2| = 2a$ and parabola; constanct distance from point and line) in standard position in Cartesian plane (foci on x-axis, center at origin)?

I aksed two simillar questions before:

Is there geometric proof for the equation of hyperbola using only constant distance from two foci definition? and Is there visual or intuitive explanation of equations of conic sections defined in traditional way?

but either proof is not synthetic or definition is different, directrix definiton (and in 2nd answer does not mention hyperbola and parabola)

Shortly: is there geometric proof of well known equations: $\frac{x^2}{a^2} \pm \frac{y^2}{b^2} = 1, y^2 = ax$ using definitions (mentioned above) of ellipse, hyperbola and parabola?

1b3b
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    If you're going to keep asking minor variations of the same question, link to the old ones. I see https://math.stackexchange.com/questions/3621086 and https://math.stackexchange.com/questions/3656767 -- any others? This will help people avoid repeating the same answers. – David K Jun 09 '20 at 12:22
  • You ask for a geometric equivalence. An equivalence between what and what? You've given only one definition of an ellipse, one for a hyperbola, and one for a parabola. What do you want to show these to be equivalent to? – David K Jun 09 '20 at 12:24
  • Geometric equivalence of Cartesian equation – 1b3b Jun 09 '20 at 12:26
  • Write that in the question and make it explicit. Show the equation(s). Then be prepared to see some algebra, because an equation is algebra. – David K Jun 09 '20 at 12:27
  • Yes some algebra (4 basic operations), not 90% of algebra. I am just asking for basic synthetic proof, is that hard? – 1b3b Jun 09 '20 at 12:31
  • @1b3b: "Geometric equivalence of Cartesian equation" ... I don't know what that means. For an ellipse, the Cartesian translation of the defining property ("sum of distances from two fixed points, say $(\pm c,0)$, is constant, say $2a$") is $$\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$$ That's arguably "the" equation for the ellipse; what we know as the "standard form" is just that equation manipulated to remove its square roots, which is algebraic window-dressing. It's not clear what kind of "synthetic proof" you're expecting here. – Blue Jun 09 '20 at 12:38
  • English is not my mother tongue, so I apologize for the possible confusion. So I'm looking for synthetic proof for the ellipse, hyperbola and parabola equations. I want a geometric argument that is equivalent to proving an equation - simply geometrically prove the equation, with minimal analytical methods. – 1b3b Jun 09 '20 at 12:41
  • This is should be quite important. Because, for example, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ implies that ellipse is streched circle. Why is that? For this kind of questions algebra is just awful – 1b3b Jun 09 '20 at 12:44
  • Do you consider the distance-formula-based version of the ellipse equation as having been "geometrically proven"? – Blue Jun 09 '20 at 12:44
  • Yes, because it is direct consequence of Pythagorean theorem. However, this procedure involves other analytical methods, so often proof is not geometric. But if this coordination can be kept to a minimum, then the proof can pass as geometric, in fact. – 1b3b Jun 09 '20 at 12:49
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    "$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$" -- is that the specific equation that you want someone to show equivalent to $r_1 + r_2 = 2a$? Write this in the question, because it is not so visible in the comments. – David K Jun 09 '20 at 13:08
  • I made a note about the position of the conics in Cartesian plane (foci on x-axis, center at origin), but I will further emphasize that – 1b3b Jun 09 '20 at 13:21
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    You wrote "I want a geometric argument that is equivalent to proving" the equations of conic sections. May I ask how would you describe geometrically those equations? Just to understand what you are looking for. – Intelligenti pauca Jun 09 '20 at 15:46
  • Well, one can treat coordinates as vertical distances or more geometrically, perpendicular to the lines of symmetry of conics and so on. So, if one prove "equation" or relationship (basically which is called equation in analytic geometry) for one point and proof can be easly done for all points (ok nobody can do $\infty$ many actions so maybe better expression is that relationship holds for all points) one proved equations of conics. – 1b3b Jun 09 '20 at 18:00
  • For clarity: relationship between those lines (coordinates) and conics (other geometrical objects in conics) in that way which simply prove equation.* – 1b3b Jun 09 '20 at 18:10
  • @Aretino, are you maybe proving this thing or is this death of my hope? – 1b3b Jun 10 '20 at 11:55
  • @1b3b I already did that in the case of a hyperbola. You probably see in my proof too many "algebraic" manipulations, but avoiding them would largely increase the length of the proof, transforming it into an ugly mess. I hope someone else can find a different proof you will like better. – Intelligenti pauca Jun 10 '20 at 20:44
  • A final remark. In my opinion this "geometric" vs "algebraic" attitude is more a matter of personal taste and education than a real thing. For instance: Apollonius of Perga, at the beginning of his treatise on conic sections (3rd century BC), geometrically derives the equivalent of what we call today the equations of parabola, ellipse and hyperbola, starting from their definitions as sections of a cone. The focal properties of ellipse and hyperbola, on the contrary, are proved only at the very end of the treatise and never used again. Would you label his treatise as "algebraic"? – Intelligenti pauca Jun 10 '20 at 20:56
  • @1b3b are you looking for this? https://en.m.wikipedia.org/wiki/Dandelin_spheres – user376343 Jun 10 '20 at 21:08
  • @Aretino yes it is personal taste. But I think it doesn’t matter if it’s my taste or not. The mathematical question is universal, no matter who asks it. Likewise, the level of education should not constrain such questions. The beauty of geometry again does not depend on new mathematical knowledge. Also, Apollonius proofs were synthetic, so my answer for your question is no. I am very sad that nobody can't or won't answer this question. Last thing: thanks for all your effort, I am sure you gave your best. If you know some modern literature or something... Greeting – 1b3b Jun 10 '20 at 21:18
  • @user376343, thanks. Dandelin spheres are beautiful tool to prove that cross sections of cone and plane are indeed ellipse, hyperbola, parabola and obviously circle. But, Dandelin's proofs don't prove equations of conics in standard position in Cartesian plane (because position obviously influences those equations). That is all I want (elementary geometric proof). – 1b3b Jun 10 '20 at 21:23
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    @1b3b You should find what you want in this book. The equation of a parabola, for instance, is proved on Prop. IX (page 11). Ellipse and hyperbola are defined via focus/directrix, but that doesn't make much difference. Enjoy! – Intelligenti pauca Jun 10 '20 at 22:35
  • @Aretino thanks! – 1b3b Jun 11 '20 at 10:33

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If you choose the construction carefully, Dandelin spheres link the standard-position formulas of the ellipse, parabola, and hyperbola almost directly to the focus-focus definitions of the ellipse and hyperbola and the focus-directrix definition of the parabola. (You can also use the same constructions to link focus-directrix definitions of the ellipse and hyperbola to their formulas.)

Here is an outline of the constructions without details of the proofs.

The Ellipse

For the ellipse, given $a > b > 0,$ let $c = \sqrt{a^2 - b^2}.$ In three-dimensional space with Cartesian $x, y, z$ coordinates, place spheres of radius $b$ at the points $F = (c,0,b)$ and $F ' = (-c,0,-b).$ Hence the spheres are tangent to the $x,y$ plane along the $x$ axis at points $F, F'$ at a distance $c$ on either side of the origin, but are on opposite sides of the $x,y$ plane (one above, one below). Note that you can construct spheres with these properties synthetically; I give the coordinates of their centers merely for reference.

An infinite cylinder of radius $b$ with axis passing through the centers of both spheres is tangent to the spheres and intersects the plane in a closed curve. Using the usual method with Dandelin spheres (which typically uses spheres of unequal size inscribed in a cone, but which works equally well for spheres of equal size inscribed in a cylinder), one can prove that this closed curve is an ellipse with semi-major axis $a$ and semi-minor axis $b$ defined by the locus of points $P$ in the $x,y$ plane such that the sum of distances $PF$ and $PF'$ is $2a$ (the focus-focus) definition.

By means of a circular cross-section of the cylinder and the use of similar triangles parallel to the $x,z$ plane, one can also show that the ellipse is a "stretched" circle with equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. $$

The Hyperbola

For the hyperbola, given $a > 0$ and $b > 0,$ let $c = \sqrt{a^2 + b^2}.$ Place spheres of radius $b$ at the points $F = (c,0,b)$ and $F ' = (-c,0,b).$ Hence the spheres are tangent to the $x,y$ plane along the $x$ axis at points $F, F'$ at a distance $c$ on either side of the origin and are on the same side of the plane.

Construct an infinite double cone with vertex at $(0,0,b)$ and axis through the centers of both spheres; make the surface of the cone tangent to the spheres. As before, you can construct the spheres and cone synthetically; the coordinates are merely for reference.

Using the usual method with Dandelin spheres, show that an point $P$ on the intersection of the cone with the $x,y$ plane satisfies $\lvert PF - PF'\rvert = 2a,$ so the intersection is a hyperbola according to the focus-focus definition.

Each circular cross-section of the cone is a circle parallel to the $y,z$ plane with a center on the line $y=0,\ z=b.$ If the circle's center is at $(x,0,b)$ then its radius is $r(x) = \frac ba x.$ So if $\lvert x\rvert < a$ the circle does not intersect the $x,y$ plane at all; but if $\lvert x\rvert = a$ the circle is tangent to the plane on the $x$ axis and if $\lvert x\rvert > a$ the circle intersects the $x,y$ plane at two points with $y$-coordinates that satisfy $$ y^2 + b^2 = (r(x))^2 = \frac{b^2}{a^2} x^2, $$ that is, $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. $$

The Parabola

For the parabola, given $a > 0,$ let $f = \frac{1}{4a}.$ Construct a sphere of radius $r_0 = \sqrt{1 - f^2}$ with center $(f,0,r_0)$ so that the sphere is tangent to the $x,y$ plane at $F = (f, 0, 0).$ Construct an infinite double cone with vertex at $(2f - 4a, 0, 2r_0)$ with its axis through the center of the sphere, and make the cone tangent to the sphere. That is, the center of the sphere is at a distance $f$ from the $z$ axis and $1$ from the origin, the distance from the $x,y$ plane to the vertex of the cone is the diameter of the sphere, and a line from the origin to the vertex is tangent to the sphere. The sphere, vertex, and cone can be constructed synthetically relative to a given "$x,y$" plane, given a segment of length $1$ and a segment of length $a.$

One can use the usual method of a Dandelin sphere to show that the cone intersects the $x,y$ plane at points $P$ such that $PF$ is equal to the distance from $P$ to the line $x=-\frac1f,\ z=0,$ that is, the intersection is a parabola defined by a focus and directrix.

If we take a plane through the origin $(0,0,0)$ perpendicular to the axis of the cone, its cross-section of the cone is a circle tangent to $(0,0,0)$ with radius $1.$ A circular cross-section of the cone that intersects the $x,y$ plane at $(x,\pm y,0)$ has radius $r(x) = 1 + \frac{x}{4a}.$ The chord between the points $(x,\pm y,0)$ on this circle cuts the perpendicular diameter into two pieces of length $2$ and $2r(x) - 2.$ Therefore half the length of the chord is $2\sqrt{r(x) - 1},$ that is, $$ y^2 = 4(r(x) - 1) = 4\left(\frac{x}{4a}\right), $$ that is, $$ x = ay^2, $$ which is the formula for a parabola opening to the right in standard position.

I chose this orientation of the parabola because the axis of the rightward opening parabola is the same as the major axis of the ellipse and hyperbola from the earlier constructions. In order to make the formula be $y = ax^2,$ simply make the sphere tangent to the $y$ axis instead of the $x$ axis.

The construction for $a < 0$ is a mirror image of the construction for $a > 0$, and other than that works exactly the same.

David K
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