Question

How do you find the points of horizontal tangency of `r=3cos2thetasectheta`?

Answer 1

`theta=+-25.92^@`

Explanation:

Let us find `(dy)/(dx)` in terms of `r` and `theta` using `x=rcostheta` and `y=rsintheta`.

As such `(dx)/(d theta)=(dr)/(d theta)costheta-rsintheta`

and `(dy)/(d theta)=(dr)/(d theta)sintheta+rcostheta`

i.e. `(dy)/(dx)=((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)`

As `r=3cos2thetasectheta`

`(dr)/(d theta)=-6sin2thetasectheta+3cos2thetasecthetatantheta`

and `(dy)/(dx)=0`, when

`sintheta(3cos2thetasecthetatantheta-6sin2thetasectheta)+rcostheta=0`

or `3cos2thetatan^2theta-12sin^2theta+rcostheta=0`

or `r=(12sin^2theta-3cos2thetatan^2theta)/(costheta)`

or `r=12sin^2thetasectheta-3cos2thetasecthetatan^2theta`

or `r=12sin^2thetasectheta-rtan^2theta`

i.e. `r=(12sin^2thetasectheta)/(1+tan^2theta)=(2tantheta)/(1+tan^2theta)*6sintheta`

= `6sin2thetasintheta`

or `3cos2thetasectheta=6sin2thetasintheta`

or `cos2theta=sin^2 2theta`

or `cos2theta=1-cos^2 2theta`

or `cos^2 2theta+cos2theta-1=0`

or `cos2theta=(-1+sqrt5)/2=0.618`

Observe that we cannot have `cos2theta=(-1-sqrt5)/`

or `2theta=+-51.83^@`

or `theta=+-25.92^@`

i.e. at `y=rsintheta=+-3cos51.83^@sec25.92^@sin25.92^@`

= `3xx0.618xx1.112xx0.437=0.9`

The graph of curve is shown below:

graph{(x^3+xy^2-3x^2+3y^2)(y^2-0.81)=0 [-5, 5, -2.5, 2.5]}