Question

An equilateral triangle is inscribed in the ellipse \(\frac{x^{2}}{3}+y^{2}=1\) such that one of the vertex of the triangle is \((0, 1)\) and one of the altitude of the triangle is along the \(y\)-axis. If A is the area of the equilateral triangle. Then \(625A^{2}\) is equal to ________.

Hint:

The other two vertices are symmetric about the y-axis, and each side from \((0,1)\) makes \(30^{\circ}\) with the y-axis.

Answer 1

Correct Answer: 2187

The altitude is the distance from \((0,1)\) to the opposite side \(y=q\), which is \(h=1-q\). For an equilateral triangle the side is \(\frac{2h}{\sqrt{3}}\), so the half-side is \(p=\frac{h}{\sqrt{3}}\).
The vertex \((p,q)=\left(\frac{1-q}{\sqrt{3}},q\right)\) lies on \(x^{2}+3y^{2}=3\): \(\frac{(1-q)^{2}}{3}+3q^{2}=3\Rightarrow 10q^{2}-2q-8=0\), so \(q=-\frac{4}{5}\) (rejecting \(q=1\)).
Then \(h=\frac{9}{5}\), and the area is \(A=\frac{h^{2}}{\sqrt{3}}=\frac{81}{25\sqrt{3}}=\frac{27\sqrt{3}}{25}\).
\(625A^{2}=625\cdot\frac{729\cdot 3}{625}=2187\).
Answer: 2187