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`x=5+3costheta , y=-2+sintheta` Find all points (if any) of horizontal and vertical tangency to the curve.

`x=5+3cos theta`


`y= -2+sin theta`


First, take the derivative of x and y with respect to theta.


`dx/(d theta) = -3sin theta`


`dy/(d theta) = cos theta`


Take note that the slope of a tangent is equal to dy/dx.


`m= dy/dx`


To determine the dy/dx of a parametric equation, apply the formula:


`dy/dx= (dy/(d theta))/(dx/(d theta))`


When the tangent line is horizontal, the slope is zero.


`0= (dy/(d theta))/(dx/(d theta))`


This occurs when `dy/(d theta)=0` and `dx/(d theta) !=0` . So setting the derivative of y equal to zero yields:


`dy/(d theta) = 0`


`cos theta = 0`


`theta_1= pi/2+2pin`


`theta_2=(3pi)/2 + 2pin`


(where n is any integer)


So the graph of the parametric equation has horizontal tangent at these values of theta.


To determine the points (x,y), plug-in the values of theta to the given parametric equation.


`theta_1 =pi/2+2pin`


`x=5+3cos(pi/2+2pin)=5+3cos(pi/2)=5+3*0=5`


`y=-2+sin(pi/2+2pin)=-2+sin(pi/2)=-2+1=-1`


`theta_2 = (3pi)/2+2pin`


`x=5+3cos((3pi)/2+2pin)=5+3cos((3pi)/2)=5+3*0=5`


`y=-2+sin((3pi)/2+2pin)=-2+sin((3pi)/2)=-2+(-1)=-3`


Therefore, the graph of the parametric equation has horizontal tangent at points (5,-1) and (5,-3).


Moreover, when the tangent line is vertical, the slope is undefined.


`u n d e f i n e d= (dy/(d theta))/(dx/(d theta))`


This occurs when `dx/(d theta)=0`  and  `dy/(d theta)!=0` . So, setting the derivative of x equal to zero yields:


`dx/(d theta) = 0`


`-3sin theta = 0`


`sin theta = 0`


`theta_1 = 2pin`


`theta_2= pi+2pin`


(where n is any integer)


So the graph of the parametric equation has vertical tangent at these values of theta.


To determine the points (x,y), plug-in the values of theta to the given parametric equation.


`theta_1=2pin`


`x=5+3cos(2pin)=5+3cos(2pi) =5+3*1=8`


`y=-2+sin(2pin)=-2+2sin(2pi)=-2+0=-2`


`theta_2=pi+2pin`


`x=5+3cos(pi+2pin)=5+3cos(pi)=5+3(-1)=2`


`y=-2+sin(pi+2pin)=-2+sin(pi)=-2+0=-2`


Therefore, the graph of the parametric equation has vertical tangent at points (8,-2) and (2,-2).

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