(1) The straight line y=kx+2 intersects with the X axis, and the Y axis intersects with point A and point B respectively.

So, b (0,2).

Point C( 1, a) is the intersection of a straight line and a hyperbola, the intersection point c is the CD⊥y axis, and the vertical foot is d,

Therefore, CD=| 1|= 1, D(0, a),

Therefore, BD = | a-2 |

The area of △BCD is 1.

1/2 * CD * BD = 1/2 * 1 * | a-2 | = 1

So | a-2 | = 2

Therefore, a 1=4 and a2=0 (irrelevant, omitted).

So, a=4.

Let the analytical formula of hyperbola be: y = k1/x.

When x= 1, y=a=4.

Therefore, k 1=4.

So the analytic formula of hyperbola is y = 4/x.

(2) the point c on the straight line y=kx+2 is (1, 4).

So, 4=k+2.

k=2

So the straight line is y=2x+2.

So, A(- 1, 0)

Let E(0, y).

CD= 1，BD=|4-2|=2，BC=√5

AB=√5

AE=√( 1+y^2),BE=|y-2|

A triangle with vertices e, a, B a and b is similar to △BCD.

So ab 2+AE 2 = be 2.

5+ 1+y^2=(y-2)^2

y=- 1/2

Or ab 2 = AE 2+be 2.

5= 1+y^2+(y-2)^2

Y 1=0, y2=2 (coincident with point B, discarded)

y=0 .

Therefore, point E(0,-1/2) or (0,0).