Algebra 2/Ellipses

In terms of Distances

The four conic sections.

In the previous lesson we studied parabolas, which arise from slicing a cone parallel to the cone edge. We saw that it had a much more useful representation in terms of points equidistant from a point and a line.

Here we study ellipses, which arise from slicing a cone at a different angle. If the slice forms an acute angle with the axis of symmetry, which is greater than the angle formed by the cone edge, then the result is an ellipse.

But as before, this description is hard to work with. Luckily, there is a proof which shows that ellipses can be re-characterized in terms of distances, just like for parabolas!

The proof of this is challenging, but the excellent math YouTube channel 3 Brown 1 Blue provides an inspiring presentation of it: 3B1B - Why slicing a cone gives an ellipse (beautiful proof).

We will simply take the result for granted here:

Let $F_{1},F_{2}$ be any two points, and let d be any positive real number.  Then the locus of points, P, such that the sum of distances to $F_{1}$ and $F_{2}$ equals d, is an ellipse.  

Various points, P, with one length shown in blue and the other in red. Their total length is constant.

How to draw an ellipse with string, two pins, paper, and a pencil.

There is a fun alternate characterization of the same idea -- a little do-it-yourself craft that you can even share with small kids, which is always fun!

1. Take a length of string. This will represent d, the total length.
2. Take two pins and a piece of paper, and pin the string into the paper at two points. Make sure that you don't mind pressing a pin into whatever is underneath the paper! The pins will represent the two focal points, $F_{1}$ and $F_{2}$.
3. Take a pencil and place it against the string, using the pencil to pull the string firmly but without pulling out the pins.
4. Place the pencil to the paper, and then sweep out a curve on the paper, always keeping the string and pencil tight.

The result is an ellipse!

The Parameters

We call each point $F_{1},F_{2}$ a "focus", the Latin plural of which is "focii".

To begin with, for simplicity, we will assume that $F_{1}(-c,0)$ and $F_{2}(c,0)$, for a positive c. Which is to say, the focii are placed on the x-axis at equal distances to the left and right.

We call $V_{l},V_{r},V_{u},V_{d}$ the vertices, which are the the left-most, right-most, up-most, and down-most points on the ellipse. In the diagram on the right, these are labeled A, B, C, D.

Because we place the focii on the x-axis, it is clear from the diagram that there are positive numbers a, b such that $V_{l}(-a,0),V_{r}(a,0),V_{u}(0,b),V(0,-b)$, respectively.

Equations

Let $P(x,y)$ be any point on the ellipse.

We now seek an equation which defines the relationship between x, y, and the parameters a, b, c, d.

First notice that if we set P to $V_{r}$ we can see that d must equal $F_{1}V_{r}+F_{2}V_{r}$.

But $F_{1}V=c+a$ and $F_{2}V=a-c$. Therefore we obtain a first useful equation

$\color {red}d=2a$

Animation of basic ellipse quantities. The first segment of animation shows f=2a. The second segment shows $a^{2}=b^{2}+c^{2}$. The third shows a random point on the ellipse.

Next consider $V_{u}$ which must also satisfy $d=F_{1}V_{u}+F_{2}V_{u}$.

But notice that the corresponding edges now form an isosceles triangle and therefore $F_{1}V={\frac {f}{2}}=a$ and $F_{2}V=a$.

Then if we consider the right triangle formed by the origin and $F_{2}$ and $V_{u}$, it has legs c and b, and hypotenuse a.