(1) Let the equation of elliptic circle be x 2/a 2+y 2/b 2 =1.

(a & gtb & gt0)

Let C(acosθ, bsinθ), then the midpoint m of OC is (0.5acosθ, 0.5bsinθ).

Let the coordinates of A and B be (x 1, y 1) and (x2, y2) respectively, and substitute the straight line AB with a slope of k into the elliptic equation, and get:

x 1^2/a^2+y 1^2/b^2= 1

x2^2/a^2+y2^2/b^2= 1

Subtract the two expressions: k = (y1-y2)/(x1-x2) =-(b/a) 2× (x1+x2)/(y1+y2) =/kloc.

M is also the midpoint of AB, so

(x 1+x2)/(y 1+y2)= 0.5 acosθ/0.5 bsinθ

That is, BSinθ/acosθ =-(b/a) 2.

Simplify:

bcosθ+asinθ=0

……①

At the same time, the slope of MF is 1, so 0.5bsinθ/(0.5acosθ-c)= 1.

Simplify:

acosθ-bsinθ=2c

……②

① The sum of squares of expression ② gives: a 2+b 2 = 4c 2.

,

And a 2-c 2 = b 2.

∴e=c/a=√ 10/5

⑵S△OAC= 1/2S parallelogram OACB=S△OAB= 15√5.

Using the formula AB = 2ab 2/(A2-C2COS α), the chord length of the elliptical focus is obtained.

α is the inclination of straight line AB.

Here, cos α = 1/2.

,

So ab = 4ab 2/(2a 2-c 2)

Distance from o to straight line AB d=c/√2

And s △ OAB =15 √ 5 =1/2ab× d.

Substituting into the above categories, it is simplified as: a 2 = 100,

b^2=60

The equation of an elliptic circle is x 2/100+y 2/60 =1.

By the way, I'll prove the formula of focal chord length of ellipse for you:

Let the equation of elliptic circle be x 2/a 2+y 2/b 2 =1.

(a & gtb & gt0)

The straight line AB passing through the focal point F 1 intersects the ellipse at two points AB, and the inclination angle is α.

Another focus is F2, which connects AF2 and BF2.

Let af 1 = m and BF1= n.

Then, according to the definition of ellipse, AF2 = 2a-m.

，BF2=2a-n

In triangle AF 1F2, it is obtained by cosine theorem.

(2a-m)^2=m^2+(2c)^2-2m(2c)cosα

Simplification: m = b 2/(a-c * cos α)

Similarly, using cosine theorem again, we can get n = b 2/(a+c * cos α).

So ab = m+n = 2ab 2/(a 2-c 2cos α)