Q.An equilateral triangle is inscribed in the parabola $y^2 = 4ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Concept: Coordinate Geometry — intersection of a line with a parabola, combined with properties of an equilateral triangle.

Step 1: Let the parabola be $y^2 = 4ax$, with vertex at $O(0,0)$. One vertex of the equilateral triangle is at $O$. Let the other two vertices be $P$ and $Q$, symmetric about the x-axis (since the parabola is symmetric). Let $P = (at^2, 2at)$ and $Q = (at^2, -2at)$.

Step 2: For $\triangle OPQ$ to be equilateral, $OP = PQ$.

$OP = \sqrt{(at^2)^2 + (2at)^2} = at\sqrt{t^2 + 4}$

$PQ = 4at$ (vertical distance between $P$ and $Q$). …

We have the parabola $y^2 = 4ax$. Its vertex is at $(0,0)$. One vertex of the equilateral triangle is fixed at this vertex. The other two vertices lie on the parabola, and because the triangle is equilateral, the two remaining vertices must be symmetric with respect to the x-axis (the axis of the parabola). Why? Because the vertex of the parabola is on the axis, and an equilateral triangle with one vertex at the origin and the other two on a symmetric curve will itself be symmetric about that axis.

So let the other two vertices be $P$ and $Q$, with coordinates:

Here $t$ is a parameter. The symmetry ensures $P$ and $Q$ are reflections across the x-axis.

Now, the triangle has vertices $O(0,0)$, $P$, and $Q$. For it to be equilateral, all sides must be equal. The side $OP$ must equal $PQ$.

1. Find $OP$:

2. Find $PQ$: Since $P$ and $Q$ have the same x-coordinate and opposite y-coordinates,

3. Set them equal:

Cancel $a$ (since $a > 0$ for a standard parabola):

Square both sides:

Since $t = 0$ would make $P$ and $Q$ coincide with the vertex (degenerate triangle), we take $t^2 = 12$, so $|t| = 2\sqrt{3}$.

4. Find the side length: …