Parametric curve (x(t),y(t)) has a horizontal tangent if its slope `dy/dx` is zero i.e when `dy/dt=0` and `dx/dt!=0`

It has a vertical tangent if its slope approaches infinity i.e it is undefined, which implies that `dx/dt=0` and `dy/dt!=0`

Given parametric equations are:

`x=4cos^2(theta) ,y=2sin(theta)`

Here the parameter is `theta`

Let's take the derivative of x and y with respect to `theta`

`dx/(d theta)=4(2cos(theta)d/(d theta)cos(theta))`

`dx/(d theta)=4(2cos(theta)(-sin(theta)))`

`dx/(d theta)=-4(2sin(theta)cos(theta))`

Use trigonometric identity: `sin(2theta)=2sin(theta)cos(theta)`

`dx/(d theta)=-4sin(2theta)`

`dy/(d theta)=2cos(theta)`

For Horizontal tangents, set the derivative of y equal to zero

`dy/(d theta)=2cos(theta)=0`

`=>cos(theta)=0`

`=>theta=pi/2,(3pi)/2`

Let's check `dx/(d theta)` for the above angles,

For `theta=pi/2`

`dx/(d theta)=-4sin(2*pi/2)=-4sin(pi)=0`

For `theta=(3pi)/2`

`dx/(d theta)=-4sin(2*(3pi)/2)=-4sin(3pi)=0`

So, there are **no horizontal tangents**.

Now for vertical tangents, set the derivative of x equal to zero,

`dx/(d theta)=-4sin(2theta)=0`

`=>sin(2theta)=0`

`=>2theta=0,pi,2pi,3pi`

`=>theta=0,pi/2,pi,(3pi)/2`

Let's check for the above angles,

For `theta=0`

`dy/(d theta)=2cos(0)=2`

For `theta=pi/2`

`dy/(d theta)=2cos(pi/2)=0`

For `theta=pi`

`dy/(d theta)=2cos(pi)=-2`

For `theta=(3pi)/2`

`dy/(d theta)=2cos((3pi)/2)=0`

So, the curve has vertical tangents at `theta=0,pi`

Now let's find the corresponding x and y coordinates by plugging `theta` in the parametric equation,

For `theta=0`

`x=4cos^2(0)=4`

`y=2sin(0)=0`

For `theta=pi`

`x=4cos^2(pi)=4`

`y=2sin(pi)=0`

So, **the given parametric curve has vertical tangent at (4,0)**.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.