Question

求椭圆标准方程第一问求解答

Answer 1

c² = 6 = a² - b², b² = a² - 6
x²/a² + y²/(a² - 6) = 1
A(-a, 0)
AM: y = (√3/2)(x + a)
令x = 0, y = √3a/2, M(0, √3a/2), AM的中点为M'(-a/2, √3a/4)
将M'点的坐标代入椭圆, 解得a² = 8, b² = 2
椭圆为x²/8 + y²/2 = 1

B(0, -√2), A(-2√2, 0)
令P(2√2cosθ, √2sinθ)
PA的方程为: (y - 0)/(√2sinθ - 0) = [x - (-2√2)]/[2√2cosθ - (-2√2)]
令x = 0, y = √2sinθ/(cosθ + 1), N(0, √2sinθ/(cosθ + 1)
PB: (y + √2)/(√2sinθ + √2) =(x - 0)/(2√2cosθ - 0)
令y = 0, x = 2√2cosθ/(sinθ + 1), Q(2√2cosθ/(sinθ + 1), 0)
|AQ| = 2√2cosθ/(sinθ + 1) + 2√2
|BN| = √2sinθ/(cosθ + 1) + √2
二者之积为[2√2cosθ/(sinθ + 1) + 2√2][√2sinθ/(cosθ + 1) + √2]
= 4(sinθ + cosθ + 1)²/[(sinθ+1)(cosθ + 1)]
= 4[sin²θ + 2sinθ(cosθ + 1) + (cosθ+1)²]/(sinθcosθ + sinθ + cosθ + 1)
= 4(sin²θ + 2sinθcosθ + 2sinθ + cos²θ + 2cosθ + 1)/(sinθcosθ + sinθ + cosθ + 1)
= 8(sinθcosθ + sinθ + cosθ + 1)/(sinθcosθ + sinθ + cosθ + 1) (利用了sin²θ + cos²θ = 1)
= 8