Let the circle be:(x-1/2)^2+(y-r)^2=r^2 The parabola:y=8x-8x^2. These two equations can be solved simultaneously and the discriminant set to 0 to obtain points of tangency. This gives us: 256r^2+32r -63=0 which, in turn, yields r = 7/16 as the only admissible root. Ajit
Yu: Translate the parabola and the circle so that the y-axis is the axis of symmetry. y= -8(x^2 - 1/4), dy/dx = -16x. x^2 + (y-r)^2 = r^2, top-half is y=sqrt(r^2-x^2)+r, dy/dx = -x/sqrt(r^2-x^2). Let -8(x^2 - 1/4) = sqrt(r^2-x^2)+r and -16x = -x/sqrt(r^2-x^2). Solve for r = 7/16.
Let the circle be:(x-1/2)^2+(y-r)^2=r^2
ReplyDeleteThe parabola:y=8x-8x^2. These two equations can be solved simultaneously and the discriminant set to 0 to obtain points of tangency. This gives us: 256r^2+32r -63=0 which, in turn, yields r = 7/16 as the only admissible root.
Ajit
Yu:
ReplyDeleteTranslate the parabola and the circle so that the y-axis is the axis of symmetry.
y= -8(x^2 - 1/4), dy/dx = -16x.
x^2 + (y-r)^2 = r^2,
top-half is y=sqrt(r^2-x^2)+r, dy/dx = -x/sqrt(r^2-x^2).
Let -8(x^2 - 1/4) = sqrt(r^2-x^2)+r
and -16x = -x/sqrt(r^2-x^2).
Solve for r = 7/16.