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...
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)`
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