已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12
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![已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12](/uploads/image/z/5255369-17-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86%E6%83%B3x2%2BY2%2BX-6Y%2BM%3D0%E4%B8%8E%E7%9B%B4%E7%BA%BFx%2B2y-3%3D0%E7%9B%B8%E4%BA%A4%E4%BA%8Ep%2Cq%E4%B8%A4%E7%82%B9%2Co%E4%B8%BA%E5%8E%9F%E7%82%B9%2C%E4%B8%94op%E5%9E%82%E7%9B%B4%E4%BA%8Eoq%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%80%BC%E7%94%B1x%2B2y-3%3D0%E5%BE%97x%3D3-2y%E4%BB%A3%E5%85%A5x2%2By2%2Bx-6y%2Bm%3D0%E5%8C%96%E7%AE%80%E5%BE%97%EF%BC%9A5y2-20y%2B12%2Bm%3D0y1%2By2%3D4%2Cy1%26%238226%3By2%3D+%2812%2Bm%29%2F5%E6%88%91%E6%83%B3%E9%97%AE%E7%9A%84%E6%98%AF%E4%BB%A5%E4%B8%8A%E6%9C%80%E5%90%8E%E4%B8%80%E6%AD%A5y1%26%238226%3By2%3D+%2812)
已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12
已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值
由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0
化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5
我想问的是以上最后一步y1•y2= (12+m)/5 是怎么来的啊
已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12
韦达定理啊,x1,x2是二元一次方程ax^2+bx+c=0的两根,x1+x2=-b/a.x1x2=c/a
所以由5y2-20y+12+m=0这个方程,知y1•y2= (12+m)/5