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Triangle Inscribing in an Ellipse

\begin{figure}\begin{center}\BoxedEPSF{EllipseTriangle.epsf scaled 1000}\end{center}\end{figure}

To inscribe an Equilateral Triangle in an Ellipse, place the top Vertex at $(0,b)$, then solve to find the $(x,y)$ coordinate of the other two Vertices.

\begin{displaymath}
\sqrt{x^2+(b-y)^2}=2x
\end{displaymath} (1)


\begin{displaymath}
x^2+(b-y)^2=4x^2
\end{displaymath} (2)


\begin{displaymath}
3x^2=(b-y)^2.
\end{displaymath} (3)

Now plugging in the equation of the Ellipse
\begin{displaymath}
{x^2\over a^2}+{y^2\over b^2}=1,
\end{displaymath} (4)

gives
\begin{displaymath}
3a^2\left({1-{y^2\over b^2}}\right)=b^2-2by+y^2
\end{displaymath} (5)


\begin{displaymath}
y^2\left({1+3{a^2\over b^2}}\right)-2by +(b^2-3a^2)=0
\end{displaymath} (6)


$\displaystyle y$ $\textstyle =$ $\displaystyle {2b-\sqrt{4b^2-4(b^2-3a^2)\left({1+3{a^2\over b^2}}\right)}\over 2\left({1+3{a^2\over b^2}}\right)}$  
  $\textstyle =$ $\displaystyle {1-\sqrt{1-\left({1-3 {a^2\over b^2}}\right)\left({1+3{a^2\over b^2}}\right)}\over 1+3 {a^2\over b^2}}\,b,$ (7)

and
\begin{displaymath}
x=\pm a\sqrt{1-{y^2\over b^2}}\,.
\end{displaymath} (8)




© 1996-9 Eric W. Weisstein
1999-05-26