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

Question

An equilateral triangle OAB is inscribed in the parabola $y^2=4 x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having $A B$ as a diameter from the origin is

Accepted Answer

${y}^{2}=4x\implies a=1$

Let $A=({t}^{2},2t)$ and $B=({t}^{2},-2t)$ due to symmetry.

Triangle $OAB$ is equilateral:

Slope of $OA=	an ({30}^{\circ })=\frac{1}{\sqrt{3}}$

$\frac{2t-0}{{t}^{2}-0}=\frac{1}{\sqrt{3}}\implies \frac{2}{t}=\frac{1}{\sqrt{3}}\implies t=2\sqrt{3}$

Coordinates:

$\begin{gathered} A=(12,4\sqrt{3}) \ B=(12,-4\sqrt{3}) \end{gathered}$

Circle with diameter $AB$ :

Center $C=(\frac{12+12}{2},\frac{4\sqrt{3}-4\sqrt{3}}{2})=(12,0)$

Radius $r=\frac{1}{2}\sqrt{(12-12{)}^{2}+(4\sqrt{3}-(-4\sqrt{3}){)}^{2}}=4\sqrt{3}$

Minimum distance from origin $O(0,0)$ :

$\begin{gathered} d=|OC-r| \ OC=\sqrt{{12}^{2}+{0}^{2}}=12 \end{gathered}$

$d=12-4\sqrt{3}=4(3-\sqrt{3})$

Solutions · 1