(2) let the equation of l be y=kx+m, and let d = √ (1+k 2) when calculating. The simultaneous equation of L and circle O and the elimination of Y have (dx) 2+2kmx+m 2-3 = 0. ∵ There is a chord length of 3, ∴ Its discriminant △ = (2km) 2-4 (D2) (m2-3) > 0, ∴ m 2 > 3D^2①。 Let it have two roots, x 1, x2. From Vieta's theorem, X 1+X2 =-2km/d 2, x 1-x2 = (m 2-3)/d 2, ∴ from the chord length calculation formula, there is 3 = d2000.
Similarly, if the equations of L and E are established at the same time, we can get that the abdominal muscles are equal to [(2 √ 2) d/(2k2+1)] √ (2k2+1-m 2).
However, if the distance from point O to L is d= 丨 m 乸 /D, ∴S△OAB=( 1/2) 乸 AB blood 乸 d = [(√ 2) m/(2k2+) is substituted into ②. The extreme points k=0 and k = 1 can be obtained by taking the derivative of K to S. It is verified that the maximum value of S△OAB is (1/2)√2 when k = 1 (which also satisfies the condition of ①).
For reference.