ao= 1

You can get ab=√5.

Triangle aob is similar to triangle abc: ab/ac=ao/ab.

ac=5

oc=4

c(4，0)

2. Let y = AX2+BX+C.

Through a (-1, 0)

b(0，2)

c(4，0)

Available y=-0.5x2+ 1.5x+2.

3,4: bc=√20 from what is known.

Cp=√20 tons

cq=t

T=√5 is obtained from cq=cp.

By t=√5

bc=√20

It is easy to get the midpoint of P in bc, so make a straight line through P perpendicular to the X axis and also perpendicular to G.

G is the midpoint of om, and the abscissa of point P is 2.

Similarly, the ordinate of point P is 1.

Then p(2, 1)

The equation of the straight line op is y = 0.5x

Combined with the equation of the second parabola, the intersection point can be (1+√5, 1/2+√5/2).

Or (1-√5, 1/2-√5/2)