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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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)`