`r=asintheta` Find the points of horizontal and vertical tangency (if any) to the polar curve.

To find a tangent line to a polar curve,`r=f(theta)` we regard `theta` as parameter and write it's parametric equations as,

`x=rcos(theta)=f(theta)cos(theta)`

`y=rsin(theta)=f(theta)sin(theta)`

We are given the polar curve `r=asin(theta)`

Now let's convert polar equation into parametric equation,

`x=asin(theta)cos(theta)`

`y=asin(theta)sin(theta)=asin^2(theta)`

Slope of the line tangent to the parametric curve is given by the derivative `dy/dx`

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

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

`dx/(d theta)=a[sin(theta)d/(d theta)cos(theta)+cos(theta)d/(d theta)sin(theta)]`

`dx/(d theta)=a[sin(theta)(-sin(theta))+cos(theta)cos(theta)]`

`dx/(d theta)=a[-sin^2(theta)+cos^2(theta)]`

`dx/(d...

To find a tangent line to a polar curve,`r=f(theta)` we regard `theta` as parameter and write it's parametric equations as,

`x=rcos(theta)=f(theta)cos(theta)`

`y=rsin(theta)=f(theta)sin(theta)`

We are given the polar curve `r=asin(theta)`

Now let's convert polar equation into parametric equation,

`x=asin(theta)cos(theta)`

`y=asin(theta)sin(theta)=asin^2(theta)`

Slope of the line tangent to the parametric curve is given by the derivative `dy/dx`

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

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

`dx/(d theta)=a[sin(theta)d/(d theta)cos(theta)+cos(theta)d/(d theta)sin(theta)]`

`dx/(d theta)=a[sin(theta)(-sin(theta))+cos(theta)cos(theta)]`

`dx/(d theta)=a[-sin^2(theta)+cos^2(theta)]`

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

use the trigonometric identity:`cos(2theta)=cos^2(theta)-sin^2(theta)`

`dx/(d theta)=acos(2theta)`

`dy/(d theta)=a(2sin(theta)d/(d theta)sin(theta))`

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

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

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

We locate horizontal tangents by finding the points where `dy/(d theta)=0` ( provided that `dx/(d theta)!=0` )

and vertical tangents at the points where `dx/(d theta)=0` ( provided that `dy/(d theta)!=0` )

Setting the derivative of x equal to zero for locating vertical tangents,

`dx/(d theta)=0`

`acos(2theta)=0`

`=>cos(2theta)=0`

`2theta=pi/2,(3pi)/2,(5pi)/2,(7pi)/2`

`=>theta=pi/4,(3pi)/4,(5pi)/4,(7pi)/4`

Let's find the corresponding radius r for the above angles,

For `theta=pi/4`

`r=asin(pi/4)=a/sqrt(2)` 

For `theta=(3pi)/4`

`r=asin((3pi)/4)=a/sqrt(2)`

For `theta=(5pi)/4`

`r=asin((5pi)/4)=-a/sqrt(2)`

For `theta=(7pi)/4`

`r=asin((7pi)/4)=-a/sqrt(2)`

Now let's set the derivative of y equal to zero for locating horizontal tangents,

`dy/(d theta)=0`

`asin(2theta)=0`

`=>sin(2theta)=0`

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

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

Now, find the corresponding radius r for above angles,

For `theta=0`

`r=asin(0)=0`

For `theta=pi/2`

`r=asin(pi/2)=a`

For `theta=pi`

`r=asin(pi)=0`

For `theta=3pi/2`

`r=asin((3pi)/2)=-a`

Note: If we plot the polar curve , its a circle and it should have two horizontal and two vertical tangents. However we got four points because it depends on a, whether it's positive or negative.

For positive value of a ,

the polar curve has horizontal tangents at `(0,0),(a,pi/2)`

and vertical tangents at `(a/sqrt(2),pi/4),(a/sqrt(2),(3pi)/4)`

For negative value of a,

the polar curve has horizontal tangents at `(0,pi),(-a,(3pi)/2)`

and vertical tangents at `(-a/sqrt(2),(5pi)/4),(-a/sqrt(2),(7pi)/4)`