Six copies of the parabola \(y=x^2\) are arranged in the plane so that each vertex is tangent to a circle, and each parabola is tangent to its two neighbours. Find the radius of the circle.

P.S. the answer is not 5/8. Also, my LaTeX isn't showing correctly even though I clicked the LaTeX icon, sorry

Guest Aug 16, 2020

edited by
Guest
Aug 16, 2020

edited by Guest Aug 16, 2020

edited by Guest Aug 16, 2020

#2**+3 **

Your LaTex actually is displaying properly now but thanks for letting me know of your difficulties.

360/6= 60 degrees

So each parabola can take up 60 degrees from the origin.

Each half parabola takes up 30 degrees.

So the line y=(tan60) must be a tangent to the top parabola which is y=x^2+R

tan 60 = sqrt3

\(y=x^2+R\\ y'=2x\\ 2x=\sqrt3\\ x=\frac{\sqrt3}{2}\\ when \;\;x=\frac{\sqrt3}{2}\\ y=\sqrt3*\frac{\sqrt3}{2}=\frac{3}{2}\\ \text{The point of contact will be }( \frac{\sqrt3}{2},\frac{3}{2})\\~\\ y=x^2+R\\ \frac{3}{2}=( \frac{\sqrt3}{2})^2+R\\ \frac{3}{2}= \frac{3}{4}+R\\ R=\frac{3}{4}\)

Here are your parabolas

LaTex:

y=x^2+R\\

y'=2x\\

2x=\sqrt3\\

x=\frac{\sqrt3}{2}\\

when \;\;x=\frac{\sqrt3}{2}\\

y=\sqrt3*\frac{\sqrt3}{2}=\frac{3}{2}\\

\text{The point of contact will be }( \frac{\sqrt3}{2},\frac{3}{2})\\~\\

y=x^2+R\\

\frac{3}{2}=( \frac{\sqrt3}{2})^2+R\\

\frac{3}{2}= \frac{3}{4}+R\\

R=\frac{3}{4}

Melody Aug 16, 2020