So the analytical formula is y =-x 2+2x+3 =-(x- 1) 2+4, so B( 1, 4).
(2) Prove that vector EB=( 1, 1) and vector EA=(3, -3). Their dot product is 0, so EA is perpendicular to EB, so AB is the diameter of the circumscribed circle of triangle ABE. (Just prove that BC is perpendicular to AB, that is, CB is tangent. )
BE=√2, EA=3√2, so Tan ∠ EAB = EB/EA = 1/3 = Tan ∠ CBE (the title should be1/3, not 13).
So ∠CBE=∠EAB, so ∠CBA=∠CBE+∠EBA=∠EAB+∠EBA=90 degrees, so CB is perpendicular to BA, which means BC is tangent.
(3) Existence, and point O is the point to be sought. At this time, tan∠DEO=OD/OE= 1/3, so ∠DEO=∠BAE, right angles are equal, so they are similar. At this time, p (0 0,0)
In addition, when the intersection D and DP are perpendicular to DE, the Y axis is perpendicular to P, and the angle DEP is constant, then the triangle DEP is similar to the triangle EAB, and then P(0,-1/3).
(4) As shown in figure 1, point E' is to the left of point F.
Ihde s1= s △ o 'a 'e' = s △ oae = OA * OE/2 = 9/2,
S2=S△O'AK=O'K*O'A/2=(3-t)^2/2
S3=S△AGA'=GH*AA'/2=y*t/2=t^2
The abscissa of f is 1.5.
Therefore, the overlapping area s = s1-S2-S3 = 9/2-(3-t) 2/2-T2 = 3t-(3/2) T2 (0 < t <: = 1.5)
As shown in fig. 2, e' is to the right of point f,
s=S△AKG=KG*O'A/2=(3-t)^2？ ( 1.5 & lt； = t & lt=3)
When t >; At 3 o'clock, s=0.
To sum it up
s=3t-(3/2)t^2(0<； t & lt= 1.5)
s=(3-t)^2？ ( 1.5 & lt； = t & lt=3)
s=0？ (s & gt3)