设M={(x,y)| xcosθ+ysinθ-2=0} N={(x,y) | x2+3y2=6},若M∩N=空集,求θ的取植范围
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设M={(x,y)| xcosθ+ysinθ-2=0} N={(x,y) | x2+3y2=6},若M∩N=空集,求θ的取植范围
设M={(x,y)| xcosθ+ysinθ-2=0} N={(x,y) | x2+3y2=6},若M∩N=空集,求θ的取植范围
设M={(x,y)| xcosθ+ysinθ-2=0} N={(x,y) | x2+3y2=6},若M∩N=空集,求θ的取植范围
当 cosθ = 0时,|sinθ| = 1,
M = {(x,y)|y = 2/sinθ}
x^2 + 3(2/sinθ)^2 = 12 + x^2 > 6,
因此,
M∩N=空集,满足要求.
此时,θ = kPI + PI/2, k为任意整数.
当 cosθ 不等于 0时,
M = {(x,y)|x = 2/cosθ - ytanθ}
[2/cosθ - ytanθ]^2 + 3y^2 - 6
= 4(secθ)^2 - 4ysecθtanθ + y^2(tanθ)^2 + 3y^2 - 6
= y^2[(tanθ)^2 + 3] - 4ysecθtanθ + 4(secθ)^2 - 6.
若要y^2[(tanθ)^2 + 3] - 4ysecθtanθ + 4(secθ)^2 - 6不等于0
则
[4secθtanθ]^2 - 4[(tanθ)^2 + 3][4(secθ)^2 - 6] < 0,
4(secθ)^2(tanθ)^2 - [(tanθ)^2 + 3][4(secθ)^2 - 6] < 0.
记u = (tanθ)^2,
则,
4u(u+1) - [u + 3][4(u+1) - 6] < 0.
4u^2 + 4u - [u + 3][4u - 2] = 4u^2 + 4u - [4u^2 + 12u - 2u - 6] < 0
-6u + 6 < 0,
(tanθ)^2 = u > 1.
-1 < tanθ < 1.
kPI - PI/4 < θ < kPI + PI/4, k 为任意整数.
综合,有,
当k为任意整数,
θ = kPI + PI/2, 或者,kPI - PI/4 < θ < kPI + PI/4时, M∩N=空集