Area enclosed by two curves, part 2

Calculus Level 5

The parabola f ( x ) = x 2 f(x)=x^2 is tangent to the graph of g ( x ) = x 4 + a x 3 + c x 2 + b x + 1 g(x) = x^4 + ax^3 + \color{#D61F06}cx^2+ bx +1 at two distinct points. Given that the m i n i m u m {\color{#69047E}{minimum}} area enclosed by these two curves is p q \frac{p}{q} , where p p and q q are coprime positive integers, find the value of p + q p+q .

Remark : try the case for c = 1 c=1 here


This problem is part of Curves... cut or touch?


The answer is 31.

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1 solution

Mas Gooddo
Nov 18, 2015

for the minimum area g(x) should be symmeyric about y axis as parabola so b= 0 =a now at point of inter section x^2=x^4+cx^2+1 as they touch this equation have pair of coinside root so c=-1 eqution reduce to x^4-2x^2+1=0 ie (x^2-1)^2=0 x=1,1,-1,-1 and requred area $x^4-x^2+1-x^2dx from x=-1 to x=1 = 16/15 p=16 ,q=15 so p+q= 15+16=31

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