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

Let the other vertices be \((\pm p, q)\). The altitude from \((0,1)\) lies on the y-axis, so each side makes \(30^{\circ}\) with it: \(p=(1-q)\tan 30^{\circ}=\frac{1-q}{\sqrt{3}}\).
On the ellipse: \(\frac{p^{2}}{3}+q^{2}=1\Rightarrow\frac{(1-q)^{2}}{9}+q^{2}=1\Rightarrow 5q^{2}-q-4=0\Rightarrow q=1\) (the given vertex) or \(q=-\frac{4}{5}\).
So \(q=-\frac{4}{5}\), \(1-q=\frac{9}{5}\), \(p=\frac{9}{5\sqrt{3}}=\frac{3\sqrt{3}}{5}\).
Side \(=2p=\frac{6\sqrt{3}}{5}\), area \(A=\frac{\sqrt{3}}{4}\cdot\frac{108}{25}=\frac{27\sqrt{3}}{25}\).
\(A^{2}=\frac{729\cdot3}{625}=\frac{2187}{625}\), so \(625A^{2}=2187\).
Answer: 2187