已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)a+b=-cotθ.ab=-/sinθ/设直线Y=KX+C,坐标(a
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![已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)a+b=-cotθ.ab=-/sinθ/设直线Y=KX+C,坐标(a](/uploads/image/z/10756377-9-7.jpg?t=%E5%B7%B2%E7%9F%A5A%28a%2Ca%5E2%29%E3%80%81B%28b%2Cb%5E2%29%28a%E2%89%A0b%29%E4%B8%A4%E7%82%B9%E7%9A%84%E5%9D%90%E6%A0%87%2C%E6%BB%A1%E8%B6%B3a%5E2sin%CE%B8%2Bacos%CE%B8%3D1%2Cb%5E2sin%CE%B8%2Bbcos%CE%B8%3D1a%5E2sin%CE%B8%2Bacos%CE%B8-1%3D0%2Cb%5E2sin%CE%B8%2Bbcos%CE%B8-1%3D0a%3D%28-cos%CE%B8%2B%E2%88%9A%28cos%5E2%CE%B8%2B4sin%CE%B8%29%2F%282sin%CE%B8%29b%3D%28-cos%CE%B8-%E2%88%9A%28cos%5E2%CE%B8%2B4sin%CE%B8%29%2F%282sin%CE%B8%29a%2Bb%3D-cot%CE%B8.ab%3D-%2Fsin%CE%B8%2F%E8%AE%BE%E7%9B%B4%E7%BA%BFY%3DKX%2BC%2C%E5%9D%90%E6%A0%87%28a)
已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)a+b=-cotθ.ab=-/sinθ/设直线Y=KX+C,坐标(a
已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1
a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0
a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)
b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)
a+b=-cotθ.ab=-/sinθ
/设直线Y=KX+C,坐标(a,a^2),B(b,b^2)代入得:
(a+b)x-Y-ab=0 / 这步没看懂,什么意思啊,求解释!~
原点(0,0)到直线距离:
=-ab/√(1+(a+b)^2)
=1
已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)a+b=-cotθ.ab=-/sinθ/设直线Y=KX+C,坐标(a
设直线Y=KX+C,坐标(a,a²),B(b,b²)代入得:
a²=Ka+C ①
b²=Kb+C
上面两式相减得:
a²-b²=K(a-b)
(a-b)(a+b)=K(a-b)
得K=a+b
代入①得a²=(a+b)a+C,得C=-ab
所以直线方程:Y=(a+b)X-ab
即(a+b)X-Y-ab=0