直线y=x+m与2x²-y²=2交于A、B两点,若以AB为直径的圆经过原点,求m的值.怎么得来的(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8
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![直线y=x+m与2x²-y²=2交于A、B两点,若以AB为直径的圆经过原点,求m的值.怎么得来的(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8](/uploads/image/z/10201162-58-2.jpg?t=%E7%9B%B4%E7%BA%BFy%3Dx%2Bm%E4%B8%8E2x%26%23178%3B-y%26%23178%3B%3D2%E4%BA%A4%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%E8%8B%A5%E4%BB%A5AB%E4%B8%BA%E7%9B%B4%E5%BE%84%E7%9A%84%E5%9C%86%E7%BB%8F%E8%BF%87%E5%8E%9F%E7%82%B9%2C%E6%B1%82m%E7%9A%84%E5%80%BC.%E6%80%8E%E4%B9%88%E5%BE%97%E6%9D%A5%E7%9A%84%28y1-y2%29%5E2%3D%5B%28x1%2Bb%29-%28x2%2Bb%29%5D%5E2%3D%28x1-x2%29%5E2%3D8b%5E2%2B8)
直线y=x+m与2x²-y²=2交于A、B两点,若以AB为直径的圆经过原点,求m的值.怎么得来的(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8
直线y=x+m与2x²-y²=2交于A、B两点,若以AB为直径的圆经过原点,求m的值.
怎么得来的(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8
直线y=x+m与2x²-y²=2交于A、B两点,若以AB为直径的圆经过原点,求m的值.怎么得来的(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8
那是把直线y=x+b代入到2x^2-y^2=2中得到:
2x^2-(x+b)^2=2
x^2-2bx-(b^2+2)=0
x1+x2=2b,x1x2=-(b^2+2)
(x1-x2)^2=(x1+x2)^2-4x1x2=8b^2+8
由于有y1=x1+b,y2=x2+b
所以有:(y1-y2)^2=[(x1+b)-(x2+b)]^2=(x1-x2)^2=8b^2+8
AB^2=(x1-x2)^2+(y1-y2)^2=16b^2+16
AB=4√(b^2+1)
所以圆的半径=2√(b^2+1)