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)