已知三角形ABC,A(-1,0),B(1,0),角ABC=45度,求点C的轨迹方程.不好意思,题目打错了,是角ACB为45度,但您是根据角ACB为45度来做的,那是对的!正确答案是x^2+(y-1)^2=2 (y>0)和x^2+(y+1)^2=2 (y0和y0 来算才对啊,解
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![已知三角形ABC,A(-1,0),B(1,0),角ABC=45度,求点C的轨迹方程.不好意思,题目打错了,是角ACB为45度,但您是根据角ACB为45度来做的,那是对的!正确答案是x^2+(y-1)^2=2 (y>0)和x^2+(y+1)^2=2 (y0和y0 来算才对啊,解](/uploads/image/z/1139498-26-8.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2CA%28-1%2C0%29%2CB%281%2C0%29%2C%E8%A7%92ABC%3D45%E5%BA%A6%2C%E6%B1%82%E7%82%B9C%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%2C%E9%A2%98%E7%9B%AE%E6%89%93%E9%94%99%E4%BA%86%2C%E6%98%AF%E8%A7%92ACB%E4%B8%BA45%E5%BA%A6%2C%E4%BD%86%E6%82%A8%E6%98%AF%E6%A0%B9%E6%8D%AE%E8%A7%92ACB%E4%B8%BA45%E5%BA%A6%E6%9D%A5%E5%81%9A%E7%9A%84%2C%E9%82%A3%E6%98%AF%E5%AF%B9%E7%9A%84%21%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E6%98%AFx%5E2%2B%28y-1%29%5E2%3D2+%28y%3E0%29%E5%92%8Cx%5E2%2B%28y%2B1%29%5E2%3D2+%28y0%E5%92%8Cy0+%E6%9D%A5%E7%AE%97%E6%89%8D%E5%AF%B9%E5%95%8A%2C%E8%A7%A3)
已知三角形ABC,A(-1,0),B(1,0),角ABC=45度,求点C的轨迹方程.不好意思,题目打错了,是角ACB为45度,但您是根据角ACB为45度来做的,那是对的!正确答案是x^2+(y-1)^2=2 (y>0)和x^2+(y+1)^2=2 (y0和y0 来算才对啊,解
已知三角形ABC,A(-1,0),B(1,0),角ABC=45度,求点C的轨迹方程.
不好意思,题目打错了,是角ACB为45度,但您是根据角ACB为45度来做的,那是对的!
正确答案是x^2+(y-1)^2=2 (y>0)和x^2+(y+1)^2=2 (y0和y0 来算才对啊,解得1-√2 < y < √2 +1 和 -√2-1 < y < √2 -1
已知三角形ABC,A(-1,0),B(1,0),角ABC=45度,求点C的轨迹方程.不好意思,题目打错了,是角ACB为45度,但您是根据角ACB为45度来做的,那是对的!正确答案是x^2+(y-1)^2=2 (y>0)和x^2+(y+1)^2=2 (y0和y0 来算才对啊,解
x取1或-1是∠C是0°应该舍去
已知三角形ABC,A(-1,0),B(1,0),角ACB=45度,求点C的轨迹方程.
解:设C(x,y),则AC斜率k1=(y-0)/(x+1)=y/(x+1),
BC斜率k2=(y-0)/(x-1)=y/(x-1),夹角45°
∴1=tan45°=(k2-k1)/(1+k2*k1)=
[2y/(x+1)(x-1)]/[1+y*y/(x+1)(x-1)]
=...
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已知三角形ABC,A(-1,0),B(1,0),角ACB=45度,求点C的轨迹方程.
解:设C(x,y),则AC斜率k1=(y-0)/(x+1)=y/(x+1),
BC斜率k2=(y-0)/(x-1)=y/(x-1),夹角45°
∴1=tan45°=(k2-k1)/(1+k2*k1)=
[2y/(x+1)(x-1)]/[1+y*y/(x+1)(x-1)]
=2y/[(x+1)(x-1)+y*y]
==2y/[x*x-1+y*y]
∴2y/[(x*x-1+y*y]=1→
2y=x^2+y^2-1→
x^2+y^2-2y=1→
x^2+(y-1)^2=2
为所求点C的轨迹方程,它的图形是以M(0,1)为圆心,√2为半径的一个圆.
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