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Elliptic hyperbolic problem in senior high school mathematics
Solution: The focus of C2 is (√ 5,0), and a asymptote equation is y=2x. According to symmetry, Yi AB is the diameter of a circle, and AB=2a.

The half focal length of ∴c 1 is c=√5, so a 2-b 2 = 5.

Let the coordinate of the intersection of C 1 and y=2x in the first quadrant be (x, 2x), and substitute it into the equation of C 1 to obtain:

x^2=(a^2b^2)/(b^2+4a^2)

②,

According to symmetry, the chord length of the straight line y=2x cut by C65438 +0 =2x√5,

From the topic: 2x√5=2a/3, so x=a/(3√5).

② ③ A 2 = 1 1b 2.

A 2 = 1 1/2 is obtained from ① ④.

b^2= 1/2