262i am getting 9Ï€32
\hspace{-16}$Min. Area of Circle Which touches the Parabolas $\mathbf{y=x^2+1}$ and $\mathbf{x=y^2+1}$
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4 Answers
1yep it is 9pi/32.
note that min. distance will be along common normal.
from symmetry it can be seen that slope of common tangent is 1 hence common normal will be of slope -1.
write any one equation (y = m(x-1) -2am - am3) and solve for point on 1 parabola , other point will just be its reflection along y=x.
we get the points (diametrically opposite) as (2/4 , 5/4) and (5/4 , 2/4)
.:. min area = 9pi/32
1Actuallly to eleaborate its a kind of modfifcation of a jee 2001-2002(some year) sum in which u are asked something like this it was a 10 mark subjective the main observation here is that both parabolas can be obtained by refection about y=x
