Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 24

The given parametric equations are ,
x=2-picos(t), y=2t-pisin(t)
The curve crosses itself for different values of t , which give the same x and y value.
So, to get the point where the curve crosses itself, let's make a table for different values of t.(Refer attached image)
From the table , we can find that the curve crosses itself at (2,0) for t=+-pi/2
The derivative dy/dx is the slope of the line tangent to the parametric graph (x(t),y(t)).
dy/dx=(dy/dt)/(dx/dt)
dx/dt=-pi(-sin(t))=pisin(t)
dy/dt=2-picos(t)
dy/dx=(2-picos(t))/(pisin(t))
At t=pi/2 , dy/dx=(2-picos(pi/2))/(pisin(pi/2))=2/pi
Equation of the tangent line can be found by the point slope form of the line,
y-0=2/pi(x-2)
y=2/pi(x-2)
At t=-pi/2 ,dy/dx=(2-picos(-pi/2))/(pisin(-pi/2))=2/(-pi)=-2/pi
Equation of the tangent line,
y-0=-2/pi(x-2)
y=-2/pi(x-2)
Equation of the tangent lines at the point where the given curve crosses itself are :
y=2/pi(x-2), y=-2/pi(x-2)