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# Consider the collection of all curves of the form y=a−bx2 that pass through the point (2,1) where a and b are positive real numbers. If the minimum area of the region bounded by y=a−bx2 and the x−axis is √A sq. units then the value of A∈N is

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Solution

## y=a−bx2 When x=2,y=1 ⇒1=a−4b⇒a=1+4b Now, Area A=2√a/b∫0(a−bx2)dx ⇒A=2[ax−bx33]√a/b0 ⇒A=2[a√a√b−b3⋅a√ab√b] ⇒A=4a3/23√b ⇒A=43⋅(1+4b)3/2√b (∵a=1+4b) Differentiating w.r.t. b, we get dAdb=43⎡⎢ ⎢ ⎢ ⎢⎣6⋅b1/2⋅(1+4b)1/2−(1+4b)3/2⋅12⋅1√bb⎤⎥ ⎥ ⎥ ⎥⎦ ⇒dAdb=23[12b√(1+4b)−(1+4b)3/2b3/2] ⇒dAdb=23√1+4bb3/2(12b−1−4b) For max/min area, dAdb=0 ⇒b=18; a=32 By first derivative test, A has b=18 as point of minima. ∴Amin=43⋅(32)3/2⋅2√2 ⇒Amin=√48 sq. units

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