Question

高二数学解析几何问题

在△ABC中，已知A（-1,0）、B（1,0），点C在AB的上方，且角ACB=45°，求点C的轨迹的方程。最好有过程，谢谢！

Answer 1

设C(x,y)
AC直线斜率k1=y/(x+1)=tg(a1)
AB直线斜率k2=y/(x-1)=tg(a2)
|a1-a2|=45°
tg(|a1-a2|)=tg45°=1
a1>漏枝如a2
(tg(a1)-tg(a2))/(1+tg(a1)tg(a2))=1
(tg(a1)-tg(a2))=1+tg(a1)tg(a2)
y/(x+1)-y/(x-1)=1+y/返启(x+1)y/(x-1)
xy-y-xy-y=x^2-1+y^2
x^2+y^2-2y-1=0
x^2+(y-1)^2=2 x>0
a1<a2
(tg(a2)-tg(a1))/(1+tg(a1)tg(a2))=1
(tg(a2)-tg(a1))=1+tg(a1)tg(a2)
y/(x-1)-y/(x+1)=1+y/(x+1)y/搭纳(x-1)
xy+y-xy+y=x^2-1+y^2
x^2+y^2+2y-1=0
x^2+(y+1)^2=2 x<0

C的轨迹的方程x^2+(y+1)^2=2 x<0 x^2+(y-1)^2=2 x>0

Answer 2

y＞0
BC直线斜率改孝缓慎野k1=y/(x+1)=tg(a1)
AC直线核模斜率k2=y/(x-1)=tg(a2)
a1-a2=45°
(tg(a1)-tg(a2))/(1+tg(a1)tg(a2))=1
(tg(a1)-tg(a2))=1+tg(a1)tg(a2)
y/(x+1)-y/(x-1)=1+y/(x+1)y/(x-1)
xy-y-xy-y=x^2-1+y^2
x^2+y^2+2y-1=0
x^2+(y+1)^2=2（y＞0）

Answer 3

我主动往你边上靠，除了暖暖，

Answer 4

y=x+1