Question

What is the equation of the tangent line of `r=6cos(theta-(2pi)/3)` at `theta=(-5pi)/3`?

Answer 1

`rcos theta=1.5`. In Cartesian form, this is x = 1.5. See the tangent-inclusive Socratic graph.

Explanation:

graph{(x^2+y^2+3x-5.2y)(x-1.64-.001y)((x-1.5)^2+(y-2.6)^2-.04)=0 [-13, 13, -6.5, 6.5]}

The equation to the family of circles through the pole r = 0 and

having center at `(a, alpha)` is `r=2acos(theta-alpha)`

So, here, the radius a = 3 and `alpha=2/3pi`.

When `theta = -5/3pi`,

`r =6cos(-7/3pi)=6(1/2)=3`.

So, the point of contact of the tangent is `P(3, -5/3pi)`.

The equation to the tangent at `P(c, beta)` is

`rsin(theta-psi)=csin(beta-psi)`, where the slope of the tangent

`m = tanpsi=(r'sintheta+rcostheta)/(r'costheta-rsintheta)`, at P

`=(r'sin(-5/3pi)+3cos(-5/3pi))/(r'cos(-5/3pi)-3sin(-5/3pi)`

`=(sqrt3/2r'+3/2)/(1/2r'-3sqrt3/2)`

`=(sqrt3r'+3)/(r'-3sqrt3)`, at P.

As `r'=-6sin(theta-2/3pi)=-6sin(-7/3pi)=3sqrt3`, at P,

`m = tan psi= oo`

`psi = 90^o`,

and for the opposite direction, this is `-90^o`.

And so, the equation to the tangent is

`rsin(theta-90^0)=3sin(-300^o-90^o)`, giving

rcostheta=1.5

In Cartesian form , this is

x=x_P=1.5.

Note that Cartesian P is

`(3cos(-5/3pi), 3 sin(-5/3pi))=(1.5, 1.5sqrt3)`