How do I calculate the length of the major axis of an ellipse? I have the eccentricity and the length of the semi-major axis.
[Math] Calculating major axis of an ellipse
conic sectionsgeometry
Related Solutions
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 $(0,-\sqrt{b^2-a^2})$.
In this answer to a related question, it is shown that $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ is simply a rotated and translated version of $$ \small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0 $$ which says the semi-major axis is $$ \left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2} $$ and the semi-minor axis is $$ \left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2} $$
Best Answer
Multiply the semi-major axis by 2, and that's the major axis.