Animated Solution for Mathematics - Conic Sections: An equilateral triangle $OAB$ is inscribed in the parabola $y^2=4x$ with the vertex $O$ at the vertex of the parabola. Then the minimum distance of the circle having $AB$ as a diameter from the origin is

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are uncovering the hidden architecture of a parabola.

Imagine standing at the origin of the coordinate plane, looking at the curve $y^2=4x$. It is a beautiful, open-ended embrace. We are tasked with inscribing an equilateral triangle $OAB$ within this curve and finding the minimum distance from the origin to a circle built upon the base of this triangle.

To conquer this, we must first define our points. The parabola $y^2=4x$ is a classic. Comparing it to the standard form $y^2=4ax$, we immediately see that $a=1$.

Any point on this parabola can be elegantly described using a parameter $t$: $A=(t^2,2t)$. This is the language of the parabola, allowing us to move from abstract geometry to concrete algebra.

Since the triangle $OAB$ is equilateral and $O$ is the origin, the symmetry of the parabola dictates that the triangle must be symmetric about the x-axis. This means point $B$ is simply the mirror image of $A$ across the x-axis. Thus, $B=(t^2,-2t)$.

Now, we invoke the defining property of our triangle: it is equilateral. This means the length of side $OA$ must equal the length of side $AB$.

The distance $OA$ from the origin $(0,0)$ to $A(t^2,2t)$ is given by the distance formula:

Next, the length of the vertical side $AB$ is the difference in the y-coordinates:

By equating these, $t\sqrt{t^2+4}=4t$, we find our master key. Since $t eq 0$, we divide by $t$ to get $\sqrt{t^2+4}=4$. Squaring both sides yields $t^2+4=16$, or simply $t^2=12$.

With $t^2=12$, we know the location of our points. The center $C$ of the circle, which has $AB$ as its diameter, is the midpoint of $AB$.

Since $A$ and $B$ share the same x-coordinate $t^2$, the midpoint is $(t^2,0)=(12,0)$. The radius $R$ is half the length of the diameter $AB$.

Since $AB=4t$ and $t=\sqrt{12}=2\sqrt{3}$, the diameter is $8\sqrt{3}$, making the radius:

Finally, we seek the minimum distance from the origin to this circle. The center $C$ is at $(12,0)$, so the distance $OC$ is $12$.

The circle extends $R$ units from the center. The closest point on the circle to the origin lies on the line segment $OC$. Therefore, the minimum distance is $d=OC-R$.

Substituting our values, we get $d=12-4\sqrt{3}$. Factoring out the $4$, we arrive at our elegant final answer: