Best Fitting Pipe in parabolic trench A work crew is digging a pipeline. The cross section of the trench is in the shape of the parabola $y = x^2$. The pipe has a circular cross section. If the pipe is too large, then the pipe will not lay on the bottom of the trench. 
(a) What is the radius of the largest pipe that will lay on the bottom of the trench?
(b) If the radius of the pipe is $3$ and the trench is in the shape of $y=ax^2$, then what is the largest value of $a$ that will make the pipe lay in the bottom of the trench?
Any tips for starting points on how to solve this problem? 
 A: (a) The trench cross-section has the equation $y=x^2$. By symmetry, the circular pipe cross-section that lies at the bottom of the trench will have its center on the $y$-axis. To reach the bottom of the trench, if the radius of the circle is $r$ then the center must be at the point $(0,r)$. You can figure out the equation for the circle from that information.
For the pipe to "lay at the bottom of the trench" there must be exactly one intersection between the two curves $y=x^2$ and the circle, namely at the origin. If there are more intersections, the pipe will not fall all the way to the bottom.
So find the intersections between the two curves, and find which values of $r$ will allow exactly one intersection. The largest such value is your desired pipe radius.
(b) Use the same idea. Find the equation for the circle of radius $3$ centered at $(0,3)$ and find the intersections with $y=ax^2$. Then find the largest value of $a$ that allows exactly one intersection.

There are other ways to solve this problem, such as using calculus to find the curvature of the parabola at its bottom. The method I gave above is suitable for pre-calculus, which is your tag for the problem.
A: Standard equation of parabola
$$ y= \frac{x^2 }{2 R} \; ,\frac{y''}{(1+y'^2)^{3/2}} =1/R$$  is the curvature that should be made to match. ( The slope at bottom $y'$ vanishes), i.e., reciprocal of radius
First case
$$ 2 R = 1;\, R =\frac12 $$
Second case is general 
comparing the coefficients, general formula $y = \dfrac{x^2}{1/a} $
$$ 1/a = 2\, R ; \, a = \frac{1}{2R} = \frac16. $$
