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 $8\sqrt{3}a.$