# On The Specifics Deriving The Equation Of Ellipse

I am trying to learn how to derive the equation of an ellipse, from this website (https://people.richland.edu/james/lecture/m116/conics/elldef.html). I am struggling, however, to prove to myself why d1 + d2 = 2a (see the diagram).

I get that when you stretch the "rope" against the major axis, you will see that d1 is equal to the length -c-(-a). But I'm struggling to prove that the same will occur when you stretch the rope on the other side of the ellipse, the side with positive c and a values. I get that by definition, the ellipse has to be symmetrical. But I want to prove that how constructing the ellipse using the "rope" would force the ellipse to be symmetrical. Can someone please show me how to do this?

We have that $$d_1+d_2=2a$$ is an assumption for the definition of the ellipse and from that by Pythagorean theorem we find the cartesian equation.
• @EthanChan A circle is defined as the curve such that the distance from the center is $d=R$. For the ellipse we assume as definition $d_1+d_2=2a$ and from here we can derive the equation. – gimusi Jun 20 '18 at 6:24
• @EthanChan It suffices that the string has fixed length $2a$. Indeed in the given figure we are assuming the focus at $(\pm c,0$ and the vertex on $x$ axis at $(a,0)$ thus at that vertex $d_1+d_2=(a-c)+(a+c)=2a$. – gimusi Jun 20 '18 at 6:32
If you consider the vertex on the left, then: $$d_1=-c-(-a)=a-c \quad\text{and}\quad d_2=c-(-a)=a+c.$$ If you consider the vertex on the right, then: $$d_1=a-(-c)=a+c \quad\text{and}\quad d_2=a-c.$$ In both cases, $d_1+d_2=2a$.
To prove the symmetry, consider any point $P$ on the ellipse and the other three points obtained by reflecting $P$ about $x$ axis, $y$ axis and origin. You can immediately verify that $d_1+d_2$ is the same for those three points as for $P$, hence those three points also belong to the ellipse.