已知圆的极坐标方程为p^2-4√2pcos(θ-π/4)+6=0,(1)将极坐标方程化为直角坐标方程 (2)设点p(x,y)在该圆上,求x+y的最值.
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![已知圆的极坐标方程为p^2-4√2pcos(θ-π/4)+6=0,(1)将极坐标方程化为直角坐标方程 (2)设点p(x,y)在该圆上,求x+y的最值.](/uploads/image/z/42700-4-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86%E7%9A%84%E6%9E%81%E5%9D%90%E6%A0%87%E6%96%B9%E7%A8%8B%E4%B8%BAp%5E2-4%E2%88%9A2pcos%28%CE%B8-%CF%80%2F4%29%2B6%3D0%2C%EF%BC%881%EF%BC%89%E5%B0%86%E6%9E%81%E5%9D%90%E6%A0%87%E6%96%B9%E7%A8%8B%E5%8C%96%E4%B8%BA%E7%9B%B4%E8%A7%92%E5%9D%90%E6%A0%87%E6%96%B9%E7%A8%8B+%EF%BC%882%EF%BC%89%E8%AE%BE%E7%82%B9p%28x%2Cy%29%E5%9C%A8%E8%AF%A5%E5%9C%86%E4%B8%8A%2C%E6%B1%82x%2By%E7%9A%84%E6%9C%80%E5%80%BC.)
已知圆的极坐标方程为p^2-4√2pcos(θ-π/4)+6=0,(1)将极坐标方程化为直角坐标方程 (2)设点p(x,y)在该圆上,求x+y的最值.
已知圆的极坐标方程为p^2-4√2pcos(θ-π/4)+6=0,
(1)将极坐标方程化为直角坐标方程
(2)设点p(x,y)在该圆上,求x+y的最值.
已知圆的极坐标方程为p^2-4√2pcos(θ-π/4)+6=0,(1)将极坐标方程化为直角坐标方程 (2)设点p(x,y)在该圆上,求x+y的最值.
(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