设A(-c,0) B(c,0) (c>0)为两动点,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)为两动点,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/1854878-14-8.jpg?t=%E8%AE%BEA%28-c%2C0%29+B%28c%2C0%29+%28c%3E0%29%E4%B8%BA%E4%B8%A4%E5%8A%A8%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%EF%BC%88a%3E0%29%2C%E6%B1%82P%E7%82%B9%E7%9A%84%E8%BD%A8%E8%BF%B9.)
设A(-c,0) B(c,0) (c>0)为两动点,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)为两动点,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)为两动点,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坐标(x,y)
(x+c)^2+y^2=a^2[(x-c)^2+y^2]
(1-a^2)x^2+2c(1+a^2)x+(1-a^2)c^2+(1-a^2)y^2=0
a=1时变成4cx=0 x=0 是y轴
a≠1时变成x^2+2c(1+a^2)/(1-a^2) *x+c^2+y^2=0
(x+c(1+a^2)/(1-a^2))^2+y^2=c^2([(1+a^2)/(1-a^2)]^2-1)
是圆