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?

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    $\begingroup$ If you're going to keep asking minor variations of the same question, link to the old ones. I see math.stackexchange.com/questions/3621086 and math.stackexchange.com/questions/3656767 -- any others? This will help people avoid repeating the same answers. $\endgroup$ – David K Jun 9 at 12:22
  • $\begingroup$ 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? $\endgroup$ – David K Jun 9 at 12:24
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    $\begingroup$ "$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$" -- 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. $\endgroup$ – David K Jun 9 at 13:08
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    $\begingroup$ 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. $\endgroup$ – Intelligenti pauca Jun 9 at 15:46
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    $\begingroup$ @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! $\endgroup$ – Intelligenti pauca Jun 10 at 22:35

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.

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