From the meaning of the question, OP = C = A 2-B 2 is 1/3b = A 2-B 2 (1).

The intersection of the tangent and the circumscribed circle is G, and the intercept coordinate on the X axis is H (-(3 √ 2)/4,0).

In Rt△BOH, ob * oh = BH * og ob = b, oh = 3 √ 2/4, BH = (b 2+9/8) (1/2), og = c = 1/3b.

When the data is brought in, it is b * 3 √ 2/4 = (B2+9/8) (1/2) *1/3b (2).

B=3 is obtained by solving.

Substituting (1) gives 2 = 10.

The elliptic equation is x 2/10+y 2/9 =1.