An equilateral triangle OAB is inscribed in the parabola y^2=4x with the vertex O at the v…

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:

Accepted Answer

Let $A=(t^2,2t),\ B=(t^2,-2t)$ by symmetry. Equilateral $\Rightarrow$ $OA$ makes $30^\circ$ with the $x$-axis: $\tan30^\circ=\dfrac{2t}{t^2}=\dfrac{1}{\sqrt3}\Rightarrow t=2\sqrt3$.

$$A=(12,4\sqrt3),\ B=(12,-4\sqrt3),\quad \text{centre}=(12,0),\ r=4\sqrt3.$$

Distance of circle from origin $=OC-r=12-4\sqrt3=4(3-\sqrt3)$.

Solutions · 1