The minimum area of circle which touches the parabolas `y=x^2+1` and `y^2=x-1` is `(9pi)/(16)s qdotu n i t` (b) `(9pi)/(32)s qdotu n i t` `(9pi)/8s qdotu n i t` (d) `(9pi)/4s qdotu n i t`

Options

A. `(9pi)/(16)` sq. unit
B. `(9pi)/(32)` sq. unit
C. `(9pi)/(8)` sq. unit
D. `(9pi)/(4)` sq. unit

Answer

B

Explanation

(2) The parabolas ` t=x^(2)+1andx=y^(2)+1` are symmetrical about y=x.
Therefore, the tangent at point A is parallel to y=x.
Therefore,
`(dy)/(dx)2xor2x=1`
`orx=(1)/(2)andy=(5)/(4)`
`:.A-=((1)/(2),(5)/(4))`
`andB-=((5)/(4))`,
`andB-=((5)/(4),(1)/(2))`
Hence, radius `=(1)/(2)sqrt(((1)/(2)-(5)/(4))+((5)/(2)-(1)/(2))^(2))`
`=(1)/(2)sqrt((9)/(16)+(9)/(16))=(3)/(8)sqrt(2)`
`:." Area"=(9pi)/(32)`