independent from the directrix, the eccentricity is defined as follows:

For a given ellipse:

  • the length of the semi-major axis = $a$

  • the length of the semi-minor = $b$

  • the distance between the foci = $2c$

  • the eccentricity is defined to be $\dfrac{c}{a}$

now the relation for eccenricity value in my textbook is $\sqrt{1- \dfrac{b^{2}}{a^{2}}}$

which I cannot prove.

  • $\begingroup$ The three quantities $a,b,c$ in a general ellipse are related. Do you know how? $\endgroup$ – Arthur Apr 14 '17 at 19:44
  • $\begingroup$ I thought I did, there's right angled triangle relation but i cant recall it $\endgroup$ – sarah Apr 14 '17 at 20:35
  • $\begingroup$ Then you should draw an ellipse, mark foci and axes, label everything $a,b$ or $c$ appropriately, and work out the relationship (working through the argument will make it a lot easier to remember the next time). Once you have that relationship, it should be able easy task to compare the two values for eccentricity. $\endgroup$ – Arthur Apr 14 '17 at 22:31
  • $\begingroup$ Have you ever try to google it? There're plenty resources in the web there!! $\endgroup$ – Ng Chung Tak Apr 15 '17 at 19:40

For two focus $A,B$ and a point $M$ on the ellipse we have the relation $MA+MB=cst$.

We can evaluate the constant at $2$ points of interest :

  • on the intersection of major axis and ellipse closest to $A$


  • on an intersection of minor axis and ellipse

we have $MA=MB$ and by pythagore $MA^2=c^2+b^2$
Combining all this gives $4a^2=(MA+MB)^2=(2MA)^2=4MA^2=4c^2+4b^2$
$\implies a^2=b^2+c^2$

Please try to solve by yourself before revealing the solution.


Consider an ellipse with center at the origin of course the foci will be at $(0,\pm{c})$ or $(\pm{c}, 0) $

As you have stated the eccentricity $e$=$\frac{c} {a}$ Note also that $c^2=a^2-b^2$, $c=\sqrt{a^2-b^2} $ where $a$ and $b$ are length of the semi major and semi minor axis and interchangeably depending on the nature of the ellipse

$e=\frac{c} {a}$ =$\frac{\sqrt{a^2-b^2}} {a}$=$\frac{\sqrt{a^2-b^2}} {\sqrt{a^2}}$

$e=\sqrt{\frac{a^2-b^2} {a^2} }$

Can you finish it from there?


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