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I don't understand how you would figure out an exact formula for the linear eccentricity (distance from the center to either focus) $c$ of an ellipse, being $c^2=a^2-b^2$, where $a$ is the length of the semi-major axis and $b$ the length of the semi-minor axis. I am imagining someone defining an ellipse as a shape where you have two foci and every point on this shape will have the same sum of distances from the foci. Now, how would you find out that every time for this shape, the $c^2=a^2-b^2$?

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  • $\begingroup$ Take a look at this picture (upload.wikimedia.org/wikipedia/commons/f/f8/…) and ask yourself why the green lines have length a. I hope this should help. If you need some more hints, just ask. $\endgroup$ – Maximilian M. Apr 30 '14 at 17:54
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    $\begingroup$ Let's start with your idea about modelling ellipses. You are completely right there. But the term "shape" is a bit hard to work with. So, let's use a set. Given a plane $E$ and two points $F_1,F_2\in E$ an ellipse is defined as the set of points $P\in E$, so that the sum of their distances to the foci is always the same (let's call this sum $s$ for the moment); $\bar{F_1P}+\bar{F_2P}=s$. If you compute this for $S_1$ in the picture you get that $s=2a$. Using that an ellipse is symmetric, you get that $\bar{F_1S_3}=\bar{F_2S_3}$ and with this the length of the green line. Then Pythagoras. $\endgroup$ – Maximilian M. Apr 30 '14 at 18:21
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As you know, the foci of an ellipse whose equation is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$are described, if $a^2>b^2$, by the coordinates $(c,0)$ and $(-c,0)$ where $c=\sqrt{a^2-b^2}$. In fact the sum of the distances of a generical point $(x,y)$ from $(c,0)$ and $(-c,0)$ is, as we can see by using the Pythagorean theorem, $$\sqrt{(x-c)^2+(y-0)^2}+\sqrt{(x-(-c))^2+(y-0)^2}$$which we can prove to be the constant $2a$ (assuming $a>0$). In fact, if we plug $c=\sqrt{a^2-b^2}$ into the equation of the ellipse, it becomes $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$$which is equivalent to $$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$which is in turn equivalent, as we can see by adding $-2a^2cx$ to both members and rearranging the addends, to $$a^2((x-c)^2+y^2)=a^2(x^2-2cx+c^2+y^2)=a^4-2a^2cx+c^2x^2=(a^2-cx)^2$$which becomes, if we calculate the square root of both members and multiplicate by $4$ $$\pm 4a\sqrt{(x-c)^2+y^2}=4(a^2-cx)$$ but, since $c=\sqrt{a^2-b^2}<a$ and $\frac{x^2}{a^2}\le 1$ (see equation of the ellipse: $\frac{y^2}{b^2}\ge 0$) and therefore, for $x>0$, $cx\le ax\le a^2$, we can chose the sign + in front of the square root:$$0=4(a^2-cx)-4a\sqrt{(x-c)^2+y^2}$$which is in turn equivalent, as we notice if we add $x^2+2cx+c^2+y^2$ to both members, to $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2-4a\sqrt{(x-c)^2+y^2}=\left(2a-\sqrt{(x-c)^2+y^2}\right)^2$$which is finally equivalent, in turn, as we see by calculating the square root of both memebrs, to$$\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}$$where we chose the positive sign for the square root $\sqrt{(x+c)^2+y^2}$ because $2a-\sqrt{(x-c)^2+y^2}\ge 0$, in fact if $2a<\sqrt{(x-c)^2+y^2}$ then $4a^2<(x-c)^2+y^2$ and $3a^2<x^2+y^2-cx-b^2$ which is impossible because if $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $a^2>b^2$ then $\frac{x^2+y^3}{a^2}<1$. The last equality means that the sum of the distances of any point from $(c,0)$ and $(-c,0)$ is $2a$, as we wanted to show.

If $b^2>a^2$ the roles of $x$ and $y$ are inverted and therefore the foci are $(0,\sqrt{b^2-a^2})$ and $(0,-\sqrt{b^2-a^2})$.

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