sδCBD =δAOB

DF, which passes through d and makes the x axis vertical, intersects the x axis at point f.

Then1/2df× BC =1/2ao× ob.

That is DF×BC=AO×OB.

BC=6，AO=4，OB=3。

∴DF=4×3/6=2

The coordinate of point d is (3/2,2)

∴OE∥DF

∴OE/DF=CO/CF

OE/2=3/(9/2)

OE=4/3, and the coordinates of point E are (0, -4/3).

Let the analytical formula be y=a(x-3)(x+3).

-4/3=a(-9)

a=4/27

y=4/27(x-3)(x+3)=4/27x^2-4/3

The analytical formula is y = 4/27x 2-4/3.