椭圆C:x2/2+y2=1,B为椭圆的上顶点,过B的两条直动线BP,BQ分别交椭圆C于点P,Q,若BP垂直BQ,求证PQ过Y轴定点...
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![椭圆C:x2/2+y2=1,B为椭圆的上顶点,过B的两条直动线BP,BQ分别交椭圆C于点P,Q,若BP垂直BQ,求证PQ过Y轴定点...](/uploads/image/z/11501427-3-7.jpg?t=%E6%A4%AD%E5%9C%86C%3Ax2%2F2%2By2%3D1%2CB%E4%B8%BA%E6%A4%AD%E5%9C%86%E7%9A%84%E4%B8%8A%E9%A1%B6%E7%82%B9%2C%E8%BF%87B%E7%9A%84%E4%B8%A4%E6%9D%A1%E7%9B%B4%E5%8A%A8%E7%BA%BFBP%2CBQ%E5%88%86%E5%88%AB%E4%BA%A4%E6%A4%AD%E5%9C%86C%E4%BA%8E%E7%82%B9P%2CQ%2C%E8%8B%A5BP%E5%9E%82%E7%9B%B4BQ%2C%E6%B1%82%E8%AF%81PQ%E8%BF%87Y%E8%BD%B4%E5%AE%9A%E7%82%B9...)
椭圆C:x2/2+y2=1,B为椭圆的上顶点,过B的两条直动线BP,BQ分别交椭圆C于点P,Q,若BP垂直BQ,求证PQ过Y轴定点...
椭圆C:x2/2+y2=1,B为椭圆的上顶点,过B的两条直动线BP,BQ分别交椭圆C于点P,Q,若BP垂直BQ,求证PQ过Y轴定点...
椭圆C:x2/2+y2=1,B为椭圆的上顶点,过B的两条直动线BP,BQ分别交椭圆C于点P,Q,若BP垂直BQ,求证PQ过Y轴定点...
x²/2+y²=1
B(0,1)
显然BP,BQ都不垂直x轴
∴设BP为y=kx+1
∵BP⊥BQ
则BQ斜率为-1/k
BQ:y=-1/kx+1
y=kx+1与x²/2+y²=1联立
得
(1+2k²)x²+4kx=0
x[(1+2k²)x+4k]=0
∵B的横坐标是0
∴P的坐标是-4k/(1+2k²)
代入y=kx+1
∴纵坐标是(1-2k²)/(1+2k²)
P(-4k/(1+2k²),(1-2k²)/(1+2k²))
y=-1/kx+1与x²/2+y²=1联立
得
(1+2/k²)x²-(4/k)x=0
B的横坐标是0
∴Q的横坐标是4k/(2+k²)
代入y=kx+1得
Q的纵坐标是(2+5k²)/(2+k²)
PQ联立
代入y=mx+b
得
3+3k²=(3+3k²)b
得到b=1
PQ恒过(0,1)