`x=2-picost , y=2t-pisint` Find the equations of the tangent lines at the point where the curve crosses itself.
The given parametric equations are ,
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)).
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,
At t=`-pi/2` ,`dy/dx=(2-picos(-pi/2))/(pisin(-pi/2))=2/(-pi)=-2/pi`
Equation of the tangent line,
Equation of the tangent lines at the point where the given curve crosses itself are :