Parabola
Equilateral Triangle in Parabola — Min Distance of Circle from Origin
nta_pyq_2026_jan
Grade 11
Question:
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
$2(8-3\sqrt{3})$
$2(3+\sqrt{3})$
$4(6+\sqrt{3})$
$4(3-\sqrt{3})$
Step-by-Step Solution
Key Concept: By symmetry: $A=(t^2,2t)$, $B=(t^2,-2t)$. $OA=t\sqrt{t^2+4}$, $AB=4t$. Equilateral: $OA=AB\Rightarrow t^2+4=16t^2/... \Rightarrow t^2=12\Rightarrow t=2\sqrt{3}$.
Min distance $=4(3-\sqrt{3})$.
Correct Answer: 4