过点M0,-1的直线L与抛物线Y=(-1/2)X^2相交于A,B,且直线OA与OB的斜率之和为1,求直线L的方程(第二种解法)第一种先设点A(X1,-X1 ^2/2) B(X2,-X2 ^2/2)根据K1+K2=1得到X1+X2=-2……①再联立 2Y1=-X1 ^2② 2Y2=-X2 ^2
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![过点M0,-1的直线L与抛物线Y=(-1/2)X^2相交于A,B,且直线OA与OB的斜率之和为1,求直线L的方程(第二种解法)第一种先设点A(X1,-X1 ^2/2) B(X2,-X2 ^2/2)根据K1+K2=1得到X1+X2=-2……①再联立 2Y1=-X1 ^2② 2Y2=-X2 ^2](/uploads/image/z/11410243-43-3.jpg?t=%E8%BF%87%E7%82%B9M0%2C-1%E7%9A%84%E7%9B%B4%E7%BA%BFL%E4%B8%8E%E6%8A%9B%E7%89%A9%E7%BA%BFY%3D%28-1%2F2%29X%5E2%E7%9B%B8%E4%BA%A4%E4%BA%8EA%2CB%2C%E4%B8%94%E7%9B%B4%E7%BA%BFOA%E4%B8%8EOB%E7%9A%84%E6%96%9C%E7%8E%87%E4%B9%8B%E5%92%8C%E4%B8%BA1%2C%E6%B1%82%E7%9B%B4%E7%BA%BFL%E7%9A%84%E6%96%B9%E7%A8%8B%28%E7%AC%AC%E4%BA%8C%E7%A7%8D%E8%A7%A3%E6%B3%95%EF%BC%89%E7%AC%AC%E4%B8%80%E7%A7%8D%E5%85%88%E8%AE%BE%E7%82%B9A%28X1%2C-X1+%5E2%2F2%29+B%EF%BC%88X2%2C-X2+%5E2%2F2%29%E6%A0%B9%E6%8D%AEK1%2BK2%3D1%E5%BE%97%E5%88%B0X1%2BX2%3D-2%E2%80%A6%E2%80%A6%E2%91%A0%E5%86%8D%E8%81%94%E7%AB%8B+2Y1%3D-X1+%5E2%E2%91%A1+2Y2%3D-X2+%5E2)
过点M0,-1的直线L与抛物线Y=(-1/2)X^2相交于A,B,且直线OA与OB的斜率之和为1,求直线L的方程(第二种解法)第一种先设点A(X1,-X1 ^2/2) B(X2,-X2 ^2/2)根据K1+K2=1得到X1+X2=-2……①再联立 2Y1=-X1 ^2② 2Y2=-X2 ^2
过点M0,-1的直线L与抛物线Y=(-1/2)X^2相交于A,B,且直线OA与OB的斜率之和为1,求直线L的方程(第二种解法)
第一种先设点A(X1,-X1 ^2/2) B(X2,-X2 ^2/2)根据K1+K2=1得到X1+X2=-2……①再联立 2Y1=-X1 ^2② 2Y2=-X2 ^2③②-③得(X1-X2)(X1+X2)+2(Y1-Y2)=0 X1+X2+2K=0 代入①得K=1∴Y=X-1第二种y=kxk=y/x=-1/2xka+kb=-1(xa+xb)=1联立L:y=kx-1,y=-1/2x²x²-2kx+2=0X1+X2=2k-1/2*2k=1k=1y=x-1以上两种都对么
过点M0,-1的直线L与抛物线Y=(-1/2)X^2相交于A,B,且直线OA与OB的斜率之和为1,求直线L的方程(第二种解法)第一种先设点A(X1,-X1 ^2/2) B(X2,-X2 ^2/2)根据K1+K2=1得到X1+X2=-2……①再联立 2Y1=-X1 ^2② 2Y2=-X2 ^2
第二种解法:设L:y=kx-1,
与y=(-1/2)x^2联立,得x^2+2kx-2=0,△=4k^2+8>0,
设A(x1,y1),B(x2,y2),则x1+x2=-2k,x1x2=-2,
OA与OB的斜率之和=y1/x1+y2/x2=(x2y1+x1y2)/(x1x2)=1,
∴x2y1+x1y2=x1x2,
由L的方程得x2(kx1-1)+x1(kx2-1)=x1x2,
∴(2k-1)x1x2-(x1+x2)=0,
∴-2(2k-1)+2k=0,2-2k=0,k=1.
奥林匹克高手告诉你高考的标准方法啦:
A(2a,,-2a^2),B(2b,-2b^2)
k(OA)+k(OB)=1
(-2a^2)/(2a)+(-2b^2)/(2b)=1
a+b=-1
k(AB)=(2a^2-2b^2)/(2b-2a)=-(a+b)=1
L:y=x-1