0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you break that down into simpler terms, because I'm still a beginner? $\endgroup$ – Ethan Chan Jun 20 '18 at 6:22
  • $\begingroup$ @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. $\endgroup$ – gimusi Jun 20 '18 at 6:24
  • $\begingroup$ Okay, but you know how an ellipse is created by stretching a string attached at 2 foci? What about stretching that string will ensure that d1+d2=2a. More specifically, that the distance between the focus point and the vertex on both sides are the same? $\endgroup$ – Ethan Chan Jun 20 '18 at 6:27
  • $\begingroup$ @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 $d_1+d_2=(a-c)+(a+c)=2a$. $\endgroup$ – gimusi Jun 20 '18 at 6:32
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.