Parabolic Petals

Let the center of the circle be the origin of the $xy$ -plane and its radius be $r$ . Then the equation of the top parabola is $y = x^2 + r$ . We note that the tangent touching two adjacent parabolas passes through the original and they are $60^\circ$ apart. The tangent in the first quadrant which the top parabola touches is $y = x \tan 60^\circ = \sqrt 3 x$ . Therefore the point where the top parabola touches the parabola on the right satisfies the following system of equation.

$\begin{cases} y = x^2 + r \\ y = \sqrt 3 x \end{cases}$ $\implies x^2+r = \sqrt 3 x$ $\implies x^2-\sqrt 3x + r = 0$ $\implies x = \dfrac {\sqrt 3 \pm \sqrt{3-4r}}2$ .

When $r$ is too small, there are two solutions. When $r$ is too big, there is no solution. When the parabola is only touching the tangent the is only one solution. This means that $3-4r = 0$ $\implies r = \frac 34$ .

Therefore, $a+b = 3+4= \boxed 7$ .

The equation of the parabola with axis as positive Y-axis is y=x^2+r, where r is the radius of the circle. y=x√3 is one tangent to this parabola. Therefore the x-coordinate of the point of contact of this tangent with the parabola is √3/2. The point of contact has the x-coordinate as a solution of the equation x^2-x√3+r=0. Comparing this with the value of x=√3/2, we obtain 3-4r=0 or r=3/4

If we consider any one of the parabola's vertex as origin and its axis as y- axis, the tangent will be at 60° and slope √3 . Lets find its equation by satisfying the common point with parabola The y- intercept of this tangent line is the center of circle. Radius = 3/4

Unfortunately, the radius can be made to be any desired value by changing the scale of the problem. Yes, at some scale, it can be $\frac34$ . The problem with the problem is that all parabola are the same relative shape. Your problem specification only said that the shape of the parabola was $y=x^2$ . It did not say that the formula of the parabola exactly was $y=x^2$ . The later statement sets the problem's scale. The first statement does not.

That is why I reported your problem.

$\text{unit}=1;\ \text{ParametricPlot}\left[\text{Evaluate}\left[\text{Join}\left[\text{Table}\left[\text{RotationTransform}\left[\frac{1}{6} (2 \pi ) i\right]\left[\frac{1}{3} (4 \text{unit}) \left\{u^2+\frac{3}{4},u\right\}\right],\left\{i,\frac{1}{2},\frac{11}{2}\right\}\right], \\ \ \ \text{unit} \left(\cos (\pi u), \sin (\pi u)\right)\right]\right],\{u,-1,1\},\text{ImageSize}\to \text{Large}\right]$

Finally, I took a chance that you meant, but didn't say, that the formula of the parabola was $y=x^2$ before rotation and translation and entered that as an answer. Fortunately for me, that worked.