过点A(-2,0)的直线交圆x²+y²=1交于P,Q两点,证明向量AP·向量AQ的值为3 求证明
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![过点A(-2,0)的直线交圆x²+y²=1交于P,Q两点,证明向量AP·向量AQ的值为3 求证明](/uploads/image/z/6854978-2-8.jpg?t=%E8%BF%87%E7%82%B9A%EF%BC%88-2%2C0%EF%BC%89%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%BA%A4%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3D1%E4%BA%A4%E4%BA%8EP%2CQ%E4%B8%A4%E7%82%B9%2C%E8%AF%81%E6%98%8E%E5%90%91%E9%87%8FAP%C2%B7%E5%90%91%E9%87%8FAQ%E7%9A%84%E5%80%BC%E4%B8%BA3+%E6%B1%82%E8%AF%81%E6%98%8E)
过点A(-2,0)的直线交圆x²+y²=1交于P,Q两点,证明向量AP·向量AQ的值为3 求证明
过点A(-2,0)的直线交圆x²+y²=1交于P,Q两点,证明向量AP·向量AQ的值为3 求证明
过点A(-2,0)的直线交圆x²+y²=1交于P,Q两点,证明向量AP·向量AQ的值为3 求证明
设圆心为O:
则:OA=(-2,0)
设过A点的直线与x轴正向的夹角为a,a∈(-π/6,π/6)
则:|OP+OQ|=4sina,OP·OQ=1*1*cos
=2(2sina)^2-1=8sina^2-1
OA与OP+OQ的夹角的余弦:sina
故:AP=OP-OA,AQ=OQ-OA
AP·AQ=(OP-OA)·(OQ-OA)
=OP·OQ+|OA|^2-OA·(OP+OQ)
=8sina^2-1+4-2*|OP+OQ|*sina
=3+8sina^2-2*4sina*sina=3
设过点A的直线l为:y=k(x+2)
y=k(x+2)
x^2+y^2=1
联立得,(1+k^2)x^2+4k^2x^2+4k^2-1=0
根据韦达定理
x1+x2=-4k^2/(1+k^2)
x1x2=(4k^2-1)/(1+k^2)
AP=(x1+2,y1),AQ=(x2+2,y2)
AP·AQ=(x1+2)(x2+2)+y1y2...
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设过点A的直线l为:y=k(x+2)
y=k(x+2)
x^2+y^2=1
联立得,(1+k^2)x^2+4k^2x^2+4k^2-1=0
根据韦达定理
x1+x2=-4k^2/(1+k^2)
x1x2=(4k^2-1)/(1+k^2)
AP=(x1+2,y1),AQ=(x2+2,y2)
AP·AQ=(x1+2)(x2+2)+y1y2=(x1+2)(x2+2)+k^2(x1+2)(x2+2)
=(1+k^2)(x1+2)(x+2)
=(1+k^2)[x1x2+2(x1+x2)+4]
=(1+k^2)[(4k^2-1)/(1+k^2)-8k^2/(1+k^2)+4]
=3
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