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 :
A) $2(8 – 3\sqrt{3})$ B) $4(6 + \sqrt{3})$ C) $4(3 – \sqrt{3})$ D) $2(3 + \sqrt{3})$
A) $2(8 – 3\sqrt{3})$ B) $4(6 + \sqrt{3})$ C) $4(3 – \sqrt{3})$ D) $2(3 + \sqrt{3})$
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Solution Steps
1
Determine the coordinates of point A and B
Let point $A$ be $(t^2, 2t)$ on the parabola $y^2 = 4x$. Vertex $O$ is at $(0,0)$. For an equilateral triangle $OAB$ symmetric about the x-axis, the angle $\angle AOx$ must be $30^\circ$ (half of $60^\circ$).
Using the slope formula: $\tan 30^\circ = \frac{2t – 0}{t^2 – 0} = \frac{2}{t}$.
$\frac{1}{\sqrt{3}} = \frac{2}{t} \implies t = 2\sqrt{3}$.
2
Calculate exact coordinates
Substituting $t = 2\sqrt{3}$ into $(t^2, 2t)$:
$x = (2\sqrt{3})^2 = 12$ and $y = 2(2\sqrt{3}) = 4\sqrt{3}$.
Thus, $A = (12, 4\sqrt{3})$ and by symmetry $B = (12, -4\sqrt{3})$.
3
Find the Center and Radius of the circle
The circle has $AB$ as a diameter. The center $C$ is the midpoint of $AB$.
Center $C = \left(\frac{12+12}{2}, \frac{4\sqrt{3}-4\sqrt{3}}{2}\right) = (12, 0)$.
Radius $R$ is half the length of $AB = \frac{4\sqrt{3} – (-4\sqrt{3})}{2} = 4\sqrt{3}$.
4
Calculate the distance from Origin to Center
The origin is $O(0,0)$. The distance to the center $C(12,0)$ is:
$OC = \sqrt{(12-0)^2 + (0-0)^2} = 12$.
5
Calculate the Minimum Distance
The minimum distance from point $O$ to the circle is $|OC – R|$.
Minimum Distance $= 12 – 4\sqrt{3} = 4(3 – \sqrt{3})$.
Minimum Distance $= 4(3 – \sqrt{3})$ (Option C)
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Theory
1. Parametric Form of a Parabola
The standard parabola $y^2 = 4ax$ allows any point on its curve to be expressed in terms of a single parameter $t$ as $(at^2, 2at)$. This representation is crucial for solving problems involving geometric shapes inscribed within the parabola, as it reduces the complexity of equations from two variables to one. For $y^2=4x$, the focal parameter $a=1$.
2. Symmetry and Equilateral Triangles
When an equilateral triangle is inscribed such that one vertex coincides with the parabola’s vertex, the parabola’s axis of symmetry becomes the triangle’s altitude or median. Since the total internal angle is $60^\circ$, each half-angle at the vertex is $30^\circ$. This symmetry implies that the y-coordinates of the other two vertices are equal in magnitude but opposite in sign.
3. Circles Defined by Diameters
If $A(x_1, y_1)$ and $B(x_2, y_2)$ are the endpoints of a diameter, the circle’s center is the midpoint $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ and the radius is $\frac{1}{2}\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. In cases where the diameter is a vertical line segment like $AB$ in this problem, the radius simplifies to half the vertical length of the chord.
4. Distance from a Point to a Circle
The shortest path from an external point to a circle lies along the line connecting the point to the circle’s center. If the distance from point $P$ to center $C$ is $d$ and the radius is $r$, the minimum distance is $|d-r|$. This principle is derived from the triangle inequality and is a standard technique for finding geometric bounds in JEE coordinate geometry.
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FAQs
1
Why is the angle $\angle AOx$ exactly $30^\circ$?
In an equilateral triangle, all angles are $60^\circ$. Because the parabola $y^2=4x$ is symmetric about the x-axis, the triangle is also symmetric, meaning the x-axis bisects the $60^\circ$ angle at vertex $O$.
2
How do we know vertex $O$ is at $(0,0)$?
The problem states $O$ is at the vertex of the parabola. For $y^2 = 4x$, the vertex is the origin $(0,0)$.
3
Can we solve this using the circle’s equation?
Yes, $(x-12)^2 + y^2 = 48$. Then find the distance from $(0,0)$ to any point on the circle and minimize it, but the center-radius method is faster.
4
What if the parabola was $x^2 = 4y$?
The logic remains the same, but symmetry would be about the y-axis, and coordinates would be swapped to $(2t, t^2)$.
5
What is the side length of the triangle?
The side length is $OA = \sqrt{12^2 + (4\sqrt{3})^2} = \sqrt{144 + 48} = \sqrt{192} = 8\sqrt{3}$.
6
Is point $O$ inside or outside the circle?
Since $OC = 12$ and $R = 4\sqrt{3} \approx 6.92$, $OC > R$, so point $O$ is outside the circle.
7
What is the maximum distance?
Maximum distance $= OC + R = 12 + 4\sqrt{3} = 4(3 + \sqrt{3})$.
8
How to calculate $4\sqrt{3}$ quickly?
$\sqrt{3} \approx 1.732$, so $4 \times 1.732 \approx 6.928$.
9
Is $AB$ a focal chord?
No, the focus is at $(1,0)$. The line $AB$ is $x=12$, which does not pass through the focus.
10
What properties of circles are tested here?
The diametric form of a circle and the concept of minimum distance from a point to a circle are the core properties tested.
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