( 1)

C is oval according to the meaning of the question.

2a=4

a=2

c=√3

b= 1

C:x？ /4+y？ = 1

(2)

L cannot coincide with the x axis, otherwise AOB will not form a triangle.

Let l:ky=(x+ 1)

X=ky- 1。

Simultaneous with elliptic equation

(k？ +4)y？ -2ky-3=0

According to Vieta's theorem.

y 1+y2=2k/(k？ +4)

y 1y2=-3/(k？ +4)

(y 1-y2)？

=(y 1+y2)？ -4y 1y 1

= 16(k？ +3)/(k？ +4)?

= 16(k？ +3)/[(k？ +3)? +2(k？ +3)+ 1]

= 16/[(k？ +3)+2+ 1/(k？ +3)]

k？ +3+ 1/(k？ +3) is a tick function in k? When +3= 1, the minimum value is taken.

But because of k? +3≥3

So (y 1-y2)?

= 16/[(k？ +3)+2+ 1/(k？ +3)]

≤ 16/[2+3+ 1/3]

=3

|y 1-y2|≤√3

Therefore, when L is perpendicular to the X axis, the area is the largest, which is 1*√3/2=√3/2.

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