# Why are two definitions of ellipses equivalent?

In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.

When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.

"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?

• I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly. Oct 29, 2013 at 10:29
• I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers? Apr 7, 2017 at 3:05
• Jan 4, 2019 at 0:38

Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.

Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(\pm c,0)$ for some $c\in[0,1)$.

By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=\sqrt{1-c^2}$. We must then prove that the equations $$\tag{1} x^2 + \left(\frac{y}{\sqrt{1-c^2}}\right)^2 = 1$$ $$\tag{2} \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2$$ are equivalent.

Rearranging (1) gives $$\tag{1'} y^2 = (1-c^2)(1-x^2)$$ and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.

Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $x\in(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.

Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.

• Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you! Jan 5, 2019 at 18:51
• @JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here. Jan 6, 2019 at 14:38
• Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do Jan 6, 2019 at 15:30
• Do you know why, given a squashed circle, do the foci lie on the major axis? It's seems so obvious, but I cannot explain it rigorously... Jun 26, 2019 at 13:38

(Stretched circle $$\Rightarrow$$ Ellipse with focis)

It's easy to show that projection of a diagram at an angle is the same as stretching. Let $$E$$ be a stretched circle, then it is not hard to find the corresponding angle s.t. $$E$$ is the projection of a circle $$C$$ at that angle.

Let $$C_0$$ be the unqiue cyclinder s.t. it contains the circle and it is normal to the plane containing the circle. Clearly $$E$$ is the intersection of $$C_0$$ with the plane containing $$E$$.

And there is a classical proof showing that the intersection of a cyclinder with a plane is an ellipse with focis, see @Element118 's answer.

(Ellipse with focis $$\Rightarrow$$ Stretched circle)

Let $$E$$ be an ellipse with focis $$F_1,F_2=(\pm c,0,0)$$, center $$O=(0,0,0)$$, major axis $$2a$$ and minor axis $$2b$$ sitting on the plane $$z=0$$. Clearly $$a^2=b^2+c^2$$.

Consider two spheres of radius $$b$$ and centers respectively $$(c,0,b)$$ and $$(-c,0,-b)$$ as the following diagram. There exists an unique cyclinder $$C_0$$ inscribing both of the spheres. Clearly $$C_0\cap \{z=0\}$$ is an ellipse $$E_1$$ with focis $$F_1$$ and $$F_2$$ due to the classical proof mentioned above. To prove that $$E_1=E$$, it reduces to showing that $$OM=a$$, which is equivalent to showing that $$\angle OAM=\angle OMA$$. Since $$\angle OAM = \angle AMJ$$, it reduces to showing $$\angle OMA=\angle AMJ$$ which is clearly true. Thus $$E_1=E$$. And we clearly have that $$E$$ is the projection of a circle, which is the same as a stretched circle. The result follows.

From 3Blue1Brown, at about 9:40, expanding the comment from @Paul Sinclair into an answer.

A squashed circle with major axis $$2a$$ and minor axis $$2b$$ can sit inside tangent to a cylinder with radius $$b$$. By constructing spheres tangent to the cylinder and the squashed circle, they would be tangent to the squashed circle at the foci, let them be $$F_1$$ and $$F_2$$.

The spheres are tangent to the cylinder, touching the cylinder at a circle each. These circles are perpendicular to the axis of the cylinder.

Pick a point $$P$$ on the perimeter of the squashed circle. $$F_1P$$ is tangent to a sphere, and another line $$l$$ from $$P$$ tangent to the sphere at $$Q_1$$ runs along parallel to the axis of the cylinder. Likewise, $$F_2P$$ is tangent to the other sphere, and there is another line $$m$$ from $$P$$ tangent to the sphere $$Q_2$$ running along, parallel to the axis of the cylinder.

Since $$l$$ and $$m$$ are parallel with common point $$P$$, they are the same line. It follows that $$F_1P+F_2P=Q_1P+Q_2P=Q_1Q_2$$. It follows that no matter what point $$P$$ we picked, $$Q_1Q_2$$ is of a constant length, which implies that $$F_1$$ and $$F_2$$ are indeed foci and the squashed circle is indeed an ellipse by the "sum of distances" definition.