Substitute e e (5,0) into parabola Y =-3/4x 2+5/4bx.

-(3/4)*25+(5/4)*5b=0

The solution is b=3.

(2)

1. So y =-3/4x 2+5/4bx.

=-(3/4)x^2+(5/4)*3x

=-(3/4)x^2+( 15/4)x

=-(3/4)(x^2-5x)

=-(3/4), that is, (4,0)

C point coordinates ((a+5)/2, a), namely: (4,3)

So BC=3, OB=4.

So oc = (3 2+4 2) (1/2) = 5.

Therefore, there is a little p in ∠COB, which makes ⊙P tangent to X axis, straight line BC and straight line OC.

Point P may be the heart of △COB or the lateral center of BC of △COB.

If point P is the kernel of △COB,

Then the radius of the inscribed circle of the triangle OCB is r = ob * BC/(Ob+OC+BC) = 4 * 3/(4+3+5) =12/1.

So the coordinate of point P is P(OB-r, r), which is P(3, 1).

If point p is the sub-center of BC side of △COB,

Then the distance from the center of BC side of OCB triangle r = ob * BC/(ob+oc-BC) = 4 * 3/(4+3-5) =12/2 = 6.

So the coordinate of point P is P(OB+R, R), which is P P( 10/0,6).

Therefore, there is a point P in ∠COB, which makes ⊙P tangent to X axis, straight line BC and straight line OC. The coordinate of point P is (3, 1) or (10,6).