Is there purely visual and intuitive approach for equations of conic sections using traditional definitions of ellipse (constant sum of distances from two foci), hyperbola (constant difference of distances from two foci) and parabola (same distances from directrise and some point)?

And, for example, why is then ellipse (we see that from equation) streched circle?? How traditional definiton imply this??

I asked for this before, and all answers were good, but they are mix of geometry and algebra. Once again, is there purely intuitive and visual proof (explanation) of this? Thanks


1 Answer 1


If you want something very visual and intuitive indeed, think of a circular disk facing a light source, so that the ray from the source to the center of the disk is perpendicular to the plane of the disk. The shadow cast on a plane is, of course, a "stretched circle." But the set of light rays intersecting the circumference of the circle forms a cone, so the shadow must be a conic section. In this sense, every conic section is a "stretched circle." If the shadow lies wholly on the plane, then it is an ellipse. Cf. my response to the question Planes cutting the conical surfaces.

Note, however, that this "stretching" is not the same as stretching the circle in the plane by scaling by a given factor in a given direction.

Here's a formal proof for the latter, based on the focus-directrix description (which is equivalent to the two-focus description) and the methods of Euclidean geometry in the plane. The key is to relate the ellipse to its auxillary circle, which is the circle whose diameter is the major axis of the ellipse. Given an ellipse with semimajor axis $a$ and semiminor axis $b$, we show that the perpendicular distance from the major axis to a point on the ellipse is $\frac{b}{a}$ times the distance from the major axis to the corresponding point on the circle.

Let $F$ be the focus of the ellipse. Let $l$ be its directrix, and let $P$ and $Q$ be its vertices. Let $C$ be an arbitrary point on the ellipse. Let $T$ and $U$ be the points where $\overleftrightarrow{CP}$ and $\overleftrightarrow{CQ}$ intersect $l$. Then $\frac{PF}{CF} = \frac{PT}{CT}$. This can be seen as follows. Let $A$ be the point where the axis of the ellipse intersects $l$, and let $B$ be the the foot of the perpendicular from $C$ to $l$ (not shown). We know that $\frac{PF}{PA} = \varepsilon = \frac{CF}{CB}$, where $\varepsilon$ is the eccentricity. On the other hand, $\triangle PAT \sim \triangle CBT$, so $\frac{PA}{CB} = \frac{PT}{CT}$. Putting these together yields the desired ratio.

$\hspace{3 cm}$enter image description here

It follows that $\overrightarrow{FT}$ bisects $\angle PFD$, where $D$ is the other endpoint of the chord from $C$ through $F$. (This is proved using the version of Euclid Proposition VI.3 for exterior angles, applied to $\triangle CFP$.) By a similar argument, $\frac{QF}{CF} = \frac{QU}{CU}$, so $\overrightarrow{FU}$ bisects $\angle PFC$. It follows that $\angle TFU$ is right. Therefore $AF$ is the geometric mean of $AT$ and $AU$, i.e., $AF^2 = AT \cdot AU$.

$\hspace{3 cm}$enter image description here

Now let $N$ be the foot of the perpendicular from $C$ to the major axis. Then $\triangle PNC \sim \triangle PAT$, so $\frac{CN}{PN} = \frac{AT}{AP}$. Similarly, $\frac{CN}{QN} = \frac{AU}{AQ}$. So \begin{equation} \frac{CN^2}{PN \cdot QN} = \frac{AF^2}{AP \cdot AQ} \end{equation} which is a fixed ratio, independent of $C$. If $C$ is on the minor axis, then this is seen to be $\frac{b^2}{a^2}$, so this must be the value in general.

Let $M$ be the point on the auxillary circle so that $C$ is between $E$ and $N$. Then the $\triangle PMQ$ is a right triangle, since it is inscribed in a semicircle, so $MN$ is the geometric mean of $PN$ and $QN$. It follows that $PN \cdot QN = MN^2$, hence $\frac{CN^2}{MN^2} = \frac{b^2}{a^2}$. So the ratio of $CN$ to $MN$ is $\frac{b}{a}$.

This proves that the ellipse can be obtained from the auxillary circle through vertical compression by the factor $\frac{b}{a}$. It follows then by a simple argument that the ellipse can also be obtained from the circle whose diameter is the minor axis through horizontal stretching by the factor $\frac{a}{b}$.

Translated into coordinates, where we consider the origin to be at the center of the ellipse, the $x$-axis along the major axis, and the $y$-axis along the minor axis, this yields the equation \begin{equation} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{equation} The fact that the area of the ellipse is $\pi ab$ is another consequence of this proof.

  • $\begingroup$ Wow, this seems to be an awesome answer. English is not my native language, so I will study your proof tomorrow because here is also late in my country :). If you have time, can you do similar thing (this Euclidian geometry proof) with parabola and hyperbola? Thanks, I really appreciate this. Also, I like that quote by Michael Atiyah! $\endgroup$
    – 1b3b
    May 3, 2020 at 19:31
  • $\begingroup$ Feel free to ask questions here. I'll think about the other sections... $\endgroup$ May 3, 2020 at 20:07
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    $\begingroup$ The proof is fine (+1). But I don't think your statement "the shadow cast on a plane is, of course, a stretched circle" is so obvious. I remember having a hard time to convince of that some people. The main reason being that the projection of the circle center is not the center of the ellipse. $\endgroup$ May 3, 2020 at 20:07
  • $\begingroup$ Sure. I'm using the term "stretched" somewhat loosely there. The stretching is not the same as the horizontal stretching in the plane. It's best proved using Dandelin spheres, as you mention. OP can consult my response to Planes cutting the conical surfaces if interested. $\endgroup$ May 3, 2020 at 20:16
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    $\begingroup$ Edited answer for clarity. $\endgroup$ May 3, 2020 at 20:28

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