Question

此题的直角坐标方程怎么求？

Answer 1

如图所示

Answer 2

ⅹ=(1-t^2)/(1+t^2)→ⅹ+ⅹ*t^2=1-t^2
→(1+ⅹ)*t^2=1-x
→t^2=(1-ⅹ)/(1+ⅹ)
y^2=16t^2/(1+t^2)^2
=16(1-ⅹ)/(1+ⅹ)*{1/[2/(1+ⅹ)]^2}
=16(1-ⅹ)/(1+x)*(1+x)^2/4
=4(1-x)(1+x)
=4(1-x^2)
故4ⅹ^2+y^2=4