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

Feb 2, 2017

$r \cos \theta = 1.5$. 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 $\left(a , \alpha\right)$ is $r = 2 a \cos \left(\theta - \alpha\right)$

So, here, the radius a = 3 and $\alpha = \frac{2}{3} \pi$.

When $\theta = - \frac{5}{3} \pi$,

$r = 6 \cos \left(- \frac{7}{3} \pi\right) = 6 \left(\frac{1}{2}\right) = 3$.

So, the point of contact of the tangent is $P \left(3 , - \frac{5}{3} \pi\right)$.

The equation to the tangent at $P \left(c , \beta\right)$ is

$r \sin \left(\theta - \psi\right) = c \sin \left(\beta - \psi\right)$, where the slope of the tangent

$m = \tan \psi = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$, at P

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

$= \frac{\frac{\sqrt{3}}{2} r ' + \frac{3}{2}}{\frac{1}{2} r ' - 3 \frac{\sqrt{3}}{2}}$

$= \frac{\sqrt{3} r ' + 3}{r ' - 3 \sqrt{3}}$, at P.

As $r ' = - 6 \sin \left(\theta - \frac{2}{3} \pi\right) = - 6 \sin \left(- \frac{7}{3} \pi\right) = 3 \sqrt{3}$, at P,

$m = \tan \psi = \infty$

$\psi = {90}^{o}$,

and for the opposite direction, this is $- {90}^{o}$.

And so, the equation to the tangent is

$r \sin \left(\theta - {90}^{0}\right) = 3 \sin \left(- {300}^{o} - {90}^{o}\right)$, giving

rcostheta=1.5

In Cartesian form , this is

x=x_P=1.5.

Note that Cartesian P is

$\left(3 \cos \left(- \frac{5}{3} \pi\right) , 3 \sin \left(- \frac{5}{3} \pi\right)\right) = \left(1.5 , 1.5 \sqrt{3}\right)$