设A(-c,0),B(c,0)(c>0)为两定点,动点P到A点的距离与到B点的距离的比为定值a(a>0),求P点的轨迹.
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![设A(-c,0),B(c,0)(c>0)为两定点,动点P到A点的距离与到B点的距离的比为定值a(a>0),求P点的轨迹.](/uploads/image/z/13914288-0-8.jpg?t=%E8%AE%BEA%28-c%2C0%29%2CB%28c%2C0%29%28c%3E0%29%E4%B8%BA%E4%B8%A4%E5%AE%9A%E7%82%B9%2C%E5%8A%A8%E7%82%B9P%E5%88%B0A%E7%82%B9%E7%9A%84%E8%B7%9D%E7%A6%BB%E4%B8%8E%E5%88%B0B%E7%82%B9%E7%9A%84%E8%B7%9D%E7%A6%BB%E7%9A%84%E6%AF%94%E4%B8%BA%E5%AE%9A%E5%80%BCa%28a%3E0%29%2C%E6%B1%82P%E7%82%B9%E7%9A%84%E8%BD%A8%E8%BF%B9.)
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设A(-c,0),B(c,0)(c>0)为两定点,动点P到A点的距离与到B点的距离的比为定值a(a>0),求P点的轨迹.
设A(-c,0),B(c,0)(c>0)为两定点,动点P到A点的距离与到B点的距离的比为定值a(a>0),求P点的轨迹.
设A(-c,0),B(c,0)(c>0)为两定点,动点P到A点的距离与到B点的距离的比为定值a(a>0),求P点的轨迹.
设P(x,y)坐标
PA = sqrt((x+c)^2 + y^2)
PB = sqrt((x-c)^2 + y^2)
PA/PB = a
则 sqrt((x+c)^2 + y^2)/sqrt((x-c)^2 + y^2) = a
整理发现
a=1时候是直线x=0
a不等于1
是个圆
x^2 - (2a^2*c+2c)*x/(a^2-1) +c^2+ y^2 =0