设切线斜率为k,方程为y-2 = k(x-0),kx - y + 2 = 0圆x²+y²=1圆心为原点,半径1原点与切线距离d等于半径d = |k*0 -0+2|/√(k² + 1) = 2/√(k² + 1) = 1k = ±1二切线关于y轴对称,下面只选一个(k =...算错了，答案可是12/13第4行错了，k = ±√3二切线关于y轴对称，下面只选一个(k = √3)计算.切线:y = √3x + 2代入椭圆方程: x²/4 + (√3x+ 2)² = 113x² + 16√3x+ 12= 0x₁ = (-8√3 + 6)/13, x₂ = (-8√3 -6)/13y₁= √3x₁ + 2, y₂ = √3x₂ + 2A(-6/5, 4/5), B(-2,0)AB² = (x₁ - x₂)² + (y₁ - y₂)²= (x₁ - x₂)² + (√3x₁ + 2 - √3x₂ - 2)²= 4(x₁ - x₂)²= 4(12/13)²AB = 24/13三角形OAB的面积 = (1/2)*|AB|*d = (1/2)*(24/13)*1= 12/13