# Finding the area of an equilateral triangle on an ellipse

The question is as follows:

Let $E$ be an ellipse with major axis length $4$ and minor axis length $2$. Inscribed an equilateral triangle $ABC$ in $E$ such that $A$ lies on the minor axis and $BC$ is parallel to the major axis. Compute the area of $\triangle ABC$.

My Work: Pardon my lack of talent with paint, but you get the idea. Points $a$ and $b$ have coordinates $(-x,y)$ and $(x,y)$ respectively. The equation for the ellipse in question is $x^2/4 + y^2 = 1$, which yields $y = \frac{\pm\sqrt{4-x^2}}{2}$ when rearranged, but the triangle is in the upper half so we can say that $y = \frac{\sqrt{4-x^2}}{2}$ for simplicity. Both $a$ and $b$ are at a distance of $\sqrt{x^2+y^2}$ from the origin, and a distance of $2x$ from each other. Since this is an equilateral triangle, we can say that $$2x = \sqrt{x^2+y^2} \hspace{5 mm} \text{and}$$

$\hspace{67 mm} \displaystyle{y = \frac{\sqrt{4-x^2}}{2}},$

a system of equations that yield readily to substitution. From these equations, we get that $x = \frac{2\sqrt{13}}{13}$ and $y = \frac{4\sqrt{3}}{26}$. From the dotted lines in the diagram, I hope it's clear that the square's area is $4xy$, and the triangle is half the square, so its area is $2xy$. Multiplying everything together yields the result $\frac{8\sqrt{39}}{169}$. This is a tiny number, and it seems the actual answer is $\frac{192\sqrt{39}}{169}$. I've double checked my work, and have no idea where I went wrong. Any ideas on where I went wrong, or on how to get the correct answer?

• Inscribed usually means the vertices are on the figure. And saying that $B$ is parallel to the major axis does not mean anything. Did you intend to write $BC$? The picture does not go well with the wording of the problem. – André Nicolas Jul 30 '14 at 21:16

That said, the area of the triangle you drew can be found in a simpler fashion. Note that the ellipse is $$x^2 + 4y^2 = 4,$$ and for the triangle you drew, we must have $\sqrt{x^2+y^2} = 2x$, or equivalently, $3x^2 = y^2$. Substituting immediately gives $x^2 = 4/13$ and $y^2 = 12/13$, from which it immediately follows that the area of this triangle is $x^2 \sqrt{3} = 4 \sqrt{3}/13$.
But for the triangle that was described, each vertex is located on the boundary of the ellipse. One vertex is located at a boundary point on the minor axis; the other two are parallel to the major axis. So without loss of generality suppose the vertices are $(0,-1)$, $(x,y)$, and $(-x,y)$, where $x, y > 0$. Then the distance condition becomes $$x^2 + (y+1)^2 = (2x)^2,$$ hence $x^2 = \frac{1}{3}(y+1)^2$ and $$\frac{1}{3}(y+1)^2 + 4y^2 = 4.$$ I leave the rest to you.
• @recursiverecursion: Alternately, take the line through $(0,-1)$ that makes a $30$ degree angle with the $y$-axis, see where it meets the ellipse. – André Nicolas Jul 30 '14 at 21:50