The second equation: If y=t is substituted into x = 4/5t 2, x = 4/5y 2 ... ②.
① ② The simultaneous solution is 5Y 4+ 16Y 2- 16 = 0.
(y^2+4)(5y^2-4)=0
The answer is y 2 = 5/4.
You can't see the θ range clearly, if 0
If it is 0