# x=sectheta , y=tantheta Find all points (if any) of horizontal and vertical tangency to the curve.

x=sec theta

y=tan theta

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

dx/(d theta) = sec theta tan theta

dy/(d theta) = sec^2 theta

Then, set each derivative equal to zero.

dx/(d theta) =0

• sec theta tan theta=0

sec theta = 0

theta={O/}  (There are...

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x=sec theta

y=tan theta

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

dx/(d theta) = sec theta tan theta

dy/(d theta) = sec^2 theta

Then, set each derivative equal to zero.

dx/(d theta) =0

• sec theta tan theta=0

sec theta = 0

theta={O/}  (There are no angles in which secant is zero.)

tan theta =0

theta = 0, pi

dy/ (d theta ) = 0

•  sec^2 theta=0

sec theta =0

theta = {O/}   (There are no angles in which secant is zero.)

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

m =dy/dx

To get 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 implies that the graph of the parametric equation will have a horizontal tangent when dy/(d theta)=0 and dx/(d theta)!=0 .

For the given parametric equation, there are no values of theta in which the dy/(d theta) is zero.

Therefore, the parametric equation has no horizontal tangent.

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 implies that the slope is undefined when dx/(d theta)=0 , but dy/(d theta)!=0 . This occurs at theta =0 and theta=pi .

Then, plug-in these values to the parametric equation to get the points (x,y).

theta_1=0

x_1=sec (0)=1

y_1=tan(0)=0

theta_2=pi

x_2=sec (pi)=-1

y_2=tan(pi)=0

Therefore, the parametric equation has vertical tangent at point (1,0) and (-1,0).

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