# Find the largest value of r so that the circle will touch the vertex of the parabola?

A circle of radius is dropped into the parabola . If too large, the circle will not fall all the way to the bottom, if is sufficiently small, the circle will touch the parabola at its vertex. Find the largest value of r so that the circle will touch the vertex of the parabola

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The radius of curvature of a parabola in the form:

y = (1/2p)x^2

is given by:

R = (p^2 + x^2)^(3/2)/ p^2

R is smallest when x is zero, which is at the vertex. The minimal R is p. This will give you the radius of the largest circle that can be dropped into the parabola and touch the vertex.

Since you haven't given the equation of the parabola, lets take an example:

y = 3x^2

3 = (1/2)p

R = p = 6

The largest circle that can be dropped into this parabola has radius 6.

• The radius of the circle is the radius of curvature of the parabola at the vertex. You haven't given a description of the parabola, so I'll assume the simplest case of y=ax^2. The equation for the radius R of curvature of a curve given by y=f(x) is:

R = sqrt( [(f'(x))^2 +1]^3 ) / f''(x)

Since f(x) = ax^2, f'(x)=2ax, and f''(x) = 2a, in this case, and the vertex occurs at x=0, then you can work out that R=1/(2a).

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