# the proof of the eccentricity of an ellipse

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.

• The three quantities $a,b,c$ in a general ellipse are related. Do you know how? – Arthur Apr 14 '17 at 19:44
• I thought I did, there's right angled triangle relation but i cant recall it – sarah Apr 14 '17 at 20:35
• 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. – Arthur Apr 14 '17 at 22:31
• Have you ever try to google it? There're plenty resources in the web there!! – 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 $A$

$MA+MB=2MA+AB=2(a-c)+2c=2a$

• 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?