Question

An equilateral triangle is inscribed in the parabola y² = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer 1

Let OAB be the equilateral triangle inscribed in parabola y² = 4ax. 
Let AB intersect the x-axis at point C.

Let OC = k 
From the equation of the given parabola, we have 
\(y^2 = 4ak\)
\(y = ±2\sqrt{ak}\)

∴The respective coordinates of points A and B are\( (k, 2\sqrt{ak}), and (k, -2\sqrt{ak})\)

\(AB = CA + CB\)
\(= 2\sqrt{ak} + 2\sqrt{ak}\)
\(= 4\sqrt{ak}\)

Since OAB is an equilateral triangle, OA² = AB²
\(∴ k^2 + (2\sqrt{ak})^2 = (4\sqrt{ak})^2\)
\(k^2 + 4ak = 16ak\)
\(k^2 = 12ak\)
\(k = 12a\)

\(∴ AB = 4\sqrt{ak} = 4\sqrt{(a \times 12a)}\)
\(= 4\sqrt{12a^2}\)
\(= 4\sqrt{(4a\times 3a)}\)
\(= 4(2)\sqrt{3}a\)
\(= 8\sqrt{3}a\)

Thus, the side of the equilateral triangle inscribed in parabola \(y^2 = 4ax\) is \(8\sqrt{3}a.\)