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
+118608 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}
