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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

 Aug 16, 2020
edited by Guest  Aug 16, 2020
edited by Guest  Aug 16, 2020
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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}

 Aug 16, 2020

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