What is the radius of the largest circle that you can drop all the way down into the parabola y equals x squared - so that it touches the origin?

Answer 1

A gem of a question.

The equation of the parabola is:

y = x2

And the general equation of a circle that intersects the parabola at the origin, and points upwards in the y direction, is:

x2 + (y-r)2 = r2

That is: x2 + y2= 2yr

Now, if the circle could have been 'dropped' into the parabola from above, then it must intersect with the parabola exclusively at the origin. If, on the other hand, the circle intersects the parabola elsewhere, it must be partially below the curve of the parabola, and hence could not nest within it.

Note that the whole situation is symmetrical about the y axis. If an intersection occurs on one side of the axis then it occurs on the other. It is sufficient, therefore, to consider the possible intersections in the first quadrant.

The equation of the circle in the first quadrant is:

x=sqrt(2yr - y2)

And the parabola:

x=sqrt(y)

Solving these simultaneously for any intersections we get:

sqrt(2yr - y2) = sqrt(y)

Therefore: 2yr - y2 = y and y(2r - y - 1) = 0

Clearly one solution to this is y = 0. The other root of the equation is given by solving

2r - y - 1 = 0

y = 2r - 1

In order for there to be no intersections other than the origin in the first quadrant, this root must be less than or equal to zero.

0 > 2r - 1

r < 0.5

Hence the radius of the biggest possible circle that will nest in y=x2 is 0.5.

(In general, by the way, for y = Ax2, the largest radius is 1/2A. This result can also be obtained by considering the second differential of the two functions.)

The smallest radius is 1/2A , at the bottom. So in this case (A=1) and therefore the smallest radius is 1/2 and any circle with a greater radius won't fit inside at the bottom.