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斜率为1的直线与双曲线2x^2-y^2=1相交于A、B两点,又AB中点的横坐标为1
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:设A、B的坐标为A(x1,y1) B(x2,y2)
因为AB中点的横坐标为1
所以(x1+x2)/2=1 即x1+x2=2①
因为直线斜率为1,故设直线方程为Y=x+b
将A、B点代入得y1=x1+b②
y2=x2+b③
②+③得:y1+y2=x1+x2+2b④
②-③得:y1-y2=x1-x2⑤
将A、B点代入双曲线得2(x1)^2-(y1)^2=1⑥
2(x2)^2-(y2)^2=1⑦
⑥-⑦得:2[(x1)^2-(x2)^2]-[(y1)^2-(y2)^2]=0
即 2(x1-x2)(x1+x2)-(y1-y2)(y1+Y2)=0⑧
将①④⑤代入⑧得2(x1-x2)(x1+x2)-(x1-x2)(x1+x2+2b)=0
2(x1-x2)2-(x1-x2)(2+2b)=0
(2-2b)(x1-x2)=0⑨
因为x1≠x2
所以b=1
直线方程为y=x+1
(2)y1+y2=x1+x2+2b=x1+x2+2=4⑩
y1-y2=x1-x2⑤
⑥+⑦得2[(x1)^2+(x2)^2]-[(y1)^2+(y2)^2]=2
化简为(y1)^2+(y2)^2=2[(x1)^2+(x2)^2]-2 ⑾
将⑤式两边平方得
(y1-y2)^2=(x1-x2)^2 展开得
(y1)^2+(y2)^2-2y1y2=(x1)^2+(x2)^2-2x1x2
将⑾式代入得
2[(x1)^2+(x2)^2]-2-2y1y2=(x1)^2+(x2)^2-2x1x2 移项得:
2y1y2=2[(x1)^2+(x2)^2]-2-[(x1)^2+(x2)^2-2x1x2]=(x1+x2)^2-2=2^2-2=2
y1y2=1 ⑿
而(y1-y2)^2=(Y1+Y2)^2-4Y1Y2
将⑤ ⑿式代入得
(y1-y2)^2=(Y1+Y2)^2-4Y1Y2=4^2-4=12
而x1-x2=y1-y2=12
由两点的距离公式知
|AB|=√(x1-x2)^2+(y1-y2)^2=12√2
因为AB中点的横坐标为1
所以(x1+x2)/2=1 即x1+x2=2①
因为直线斜率为1,故设直线方程为Y=x+b
将A、B点代入得y1=x1+b②
y2=x2+b③
②+③得:y1+y2=x1+x2+2b④
②-③得:y1-y2=x1-x2⑤
将A、B点代入双曲线得2(x1)^2-(y1)^2=1⑥
2(x2)^2-(y2)^2=1⑦
⑥-⑦得:2[(x1)^2-(x2)^2]-[(y1)^2-(y2)^2]=0
即 2(x1-x2)(x1+x2)-(y1-y2)(y1+Y2)=0⑧
将①④⑤代入⑧得2(x1-x2)(x1+x2)-(x1-x2)(x1+x2+2b)=0
2(x1-x2)2-(x1-x2)(2+2b)=0
(2-2b)(x1-x2)=0⑨
因为x1≠x2
所以b=1
直线方程为y=x+1
(2)y1+y2=x1+x2+2b=x1+x2+2=4⑩
y1-y2=x1-x2⑤
⑥+⑦得2[(x1)^2+(x2)^2]-[(y1)^2+(y2)^2]=2
化简为(y1)^2+(y2)^2=2[(x1)^2+(x2)^2]-2 ⑾
将⑤式两边平方得
(y1-y2)^2=(x1-x2)^2 展开得
(y1)^2+(y2)^2-2y1y2=(x1)^2+(x2)^2-2x1x2
将⑾式代入得
2[(x1)^2+(x2)^2]-2-2y1y2=(x1)^2+(x2)^2-2x1x2 移项得:
2y1y2=2[(x1)^2+(x2)^2]-2-[(x1)^2+(x2)^2-2x1x2]=(x1+x2)^2-2=2^2-2=2
y1y2=1 ⑿
而(y1-y2)^2=(Y1+Y2)^2-4Y1Y2
将⑤ ⑿式代入得
(y1-y2)^2=(Y1+Y2)^2-4Y1Y2=4^2-4=12
而x1-x2=y1-y2=12
由两点的距离公式知
|AB|=√(x1-x2)^2+(y1-y2)^2=12√2
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