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`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 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)`

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