# Find the major axis of an ellipse

1. Dec 2, 2009

### dE_logics

I want to figure out the 'a' of an ellipse (i.e. (major axis)/2) by knowledge of it's circumference and length of minor axis.

Using my little knowledge which I gained from reading (roughly) the ellipse article of wikipedia; I realized that I need to use that notorious and approximate circumference formula...so I made an equation to derive the 'a' or half of the major axis which needs to be solved.

quickmath.com gave 4 results so as to turn my vertical scroll bar into a tiny line (try it yourself, I'll post the equation).

Axiom suggest a syntax error which I know it not true, I think it has given up.

This is the equation -

h=((22/7)*(x+(c/2))*(1+(((3*((x-(c/2))/(x+(c/2)))^2))/(10+(4-(3*((x-(c/2))/(x+(c/2)))^2))^(1/2)))))/2

Which obviously I've converted from human readable format.

You need to solve for x to get the major axis.

2. Dec 2, 2009

### dE_logics

Ok, axiom has given a result -

$$\left[ {x= \%x3}, \: {x= \%x4}, \: {x={{{\sqrt {{-{{284592} \ { \%x4 \sp 2}}+{{\left( -{{189728} \ \%x3}+{{185416} \ h} -{{196504} \ c} \right)} \ \%x4} -{{284592} \ { \%x3 \sp 2}}+{{\left( {{185416} \ h} -{{196504} \ c} \right)} \ \%x3}+{{49} \ {h \sp 2}}+{{174482} \ c \ h} -{{182831} \ {c \sp 2}}}}} -{{308} \ \%x4} -{{308} \ \%x3}+{{301} \ h} -{{319} \ c}} \over {616}}}, \: {x={{-{\sqrt {{-{{284592} \ { \%x4 \sp 2}}+{{\left( -{{189728} \ \%x3}+{{185416} \ h} -{{196504} \ c} \right)} \ \%x4} -{{284592} \ { \%x3 \sp 2}}+{{\left( {{185416} \ h} -{{196504} \ c} \right)} \ \%x3}+{{49} \ {h \sp 2}}+{{174482} \ c \ h} -{{182831} \ {c \sp 2}}}}} -{{308} \ \%x4} -{{308} \ \%x3}+{{301} \ h} -{{319} \ c}} \over {616}}} \right]$$

What is this %x3?

Sounds like a substitution of x...but that's very unlikely.

3. Dec 3, 2009

### dE_logics

I computed the result (exact result...no variables) using goal seek in openoffice....and I think it's correct.

If the length of the major axis is 100, the circumference is 200, and minor axis is 0, then the minor axis comes out to be 100...which is the right answer.