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Find the equation of the line that passes trough point (0,1) such that the area, S, of the region enclosed by this line and the parabola y=x^2 is (5√5)/6.
This is what I did:
I found the equation of the line which is y=x+1
Then, I put x^2=x+1
x^2-x-1=0
α+β=-1
αβ=-1
S=integral of (x+1-x^2)dx [α,β].
It gave me (5√5)/6.
I don't know how to continue to find out the equation.
Please help me!
How did you manage to get y = x + 1? There are an INFINITE number of lines that this could possibly be. It should be y = mx + 1.
This is what I did:
I found the equation of the line which is y=mx+1
Then, I put x^2=mx+1
x^2-mx-1=0
α+β=m
αβ=-1
S=integral of (mx+1-x^2)dx [α,β].
It gave me 1/6(m+2)^3
I put 1/6(m+2)^3=(5√5)/6
It gave me that m=-2cubic root of 5√5
So, the equation is y=(=-2cubic root of 5√5)x+1
I know that my answer is wrong... because it cannot give a number like this.. but I cannot find out where is my mistake.
Attachment 26401Please have the complete solution
