求曲线x^2+4y^2=1绕原点逆时针π/3所得的曲线方程
问题描述：
求曲线x^2+4y^2=1绕原点逆时针π/3所得的曲线方程
答
变成极坐标
x=rcos(θ),y=rsin(θ)
r^2cos^2(θ) 4r^2sin^2(θ)=1
旋转π/3即
r^2cos^2(θ π/3) 4r^2sin^2(θ π/3)=1
有公式
cos(a b)=cos(a)cos(b)-sin(a)sin(b)
sin(a b)=sin(a)cos(b) cos(a)sin(b)
带回原式
r^2[cos(θ)cos(π/3)-sin(θ)sin(π/3)]^2 4r^2[sin(θ)cos(π/3) cos(θ)sin(π/3)]^2=1
[cos(π/3)x sin(π/3)y]^2 4[cos(π/3)y sin(π/3)x]^2=1
已知cos(π/3)=√3/2,sin(π/3)=1/2,
拆开化简得7x^2 13y^2 5√3xy=4
我是用手机打的,出了点问题,所有的空格都看成是加号好了