Question

图片上的题，第二问不会，求解

Answer 1

(1) p² - 4√2pcos(θ-π/4) + 6 = 0
p² - 4√2p [cosθcos(π/4) + sinθsin(π/4)] + 6 = 0 (利用两角差的馀弦公式)
p² - 4√2p [cosθ (1/√2) + sinθ (1/√2)] + 6 = 0
p² - 4pcosθ - 4psinθ + 6 = 0
x² + y² - 4x - 4y + 6 = 0 (x = pcosθ, y = psinθ, x² + y² = p²)
(x - 2)² + (y - 2)² = 2

(2) 设圆的参数方程为 x = 2 + √2cosα, y = 2 + √2sinα (设α是为免与(1)式中的θ混为一谈)
则 x + y = 2+ √2cosα + 2 + √2sinα
= 4 + √2(cosα + sinα)
= 4 + √2 * √2 [cosα (1/√2) + sinα (1/√2)]
= 4 + 2 [cosα sin(π/4) + sinα cos(π/4)]
= 4 + 2sin(α + π/4) (利用两角和的正弦公式)
当sin(α + π/4) = 1时, x + y 有最大值为6
当sin(α + π/4) = -1时, x + y 有最小值为2

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