Question

高三数学直线与椭圆的题，求详细的解答，本人基础比较差。越详细越好。

Answer 1

(1) k = 0, 直线方程为y = b (0 < b < 1)
代入椭圆x^2/4 + y^2 = 1
x^2/4 + b^2 =1
x^2 = 4(1-b^2)
x = ±2√(1-b^2)
A(2√(1-b^2), b)
B(-2√(1-b^2), 0)
|AB|=2√(1-b^2) - (-2√(1-b^2)) = 4√(1-b^2)
AB与x轴平行, 三角形AOB高为b
S = (1/2)*4√(1-b^2)*b = 2b√(1-b^2)
S' = 2√(1-b^2) + 2b*(1-b^2)^(-1/2)*(-2b)
= 2√(1-b^2) -2b^2(*1-b^2)^(-1/2) = 0
√(1-b^2) = b^2(*1-b^2)^(-1/2)
1 - b^2 = b^2
b^2 = 1/2
b = √2/2 (负值舍去)

(2)
直线方程y = kx + b代入椭圆x^2/4 + y^2 = 1
x^2/4 + (kx+b)^2 = 1
x^2 + 4(k^2*x^2 + 2kbx + b^2) = 4
(4k^2+1)x^2 + 8kbx + 4(b^2-1) = 0
x1 = [-4kb+2√(4k^2-b^2+1)]/(4k^2 + 1)
x2 = [-4kb-2√(4k^2-b^2+1)]/(4k^2 + 1)
A(x1, kx1 + b)
B(x2, kx2 + b)
|AB|^2 = (x2 - x1)^2 + k^2(x2-x1)^2 = [4√(4k^2-b^2+1)]^2 + k^2*[4√(4k^2-b^2+1)]^2
= 16(k^2+1)(4k^2-b^2+1= 2^2 = 4 (1)
底AB上的高h为O到AB的距离
S = 1 = (1/2)|AB|*h = (1/2)*2h = h
h = 1
y = kx + b, kx -y + b = 0
O(0, 0)到AB的距离: h = |k*0 -0 + b|/√(k^2 + 1) = |b|/√(k^2 + 1) = 1
b^2 = k^2 + 1 (2)
由(1)(2), k = (-1± √5)/2
k = (-1 + √5)/2 : b = ±√[(5-√5)/2], y = (-1 + √5)x/2 ±√[(5-√5)/2]
k = (-1 - √5)/2 : b = ±√[(5+√5)/2], y = (-1 - √5)x/2 ±√[(5+√5)/2]