`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).**