Thread: Finding tangency points of a polar equation

1. Finding tangency points of a polar equation

Question: Find all the points where the tangent line is horizontal and vertical for the polar equation $r=1-2cos{\theta}$

I'm trying to find the tangencies by passing through the cartesian.

So far I have this:
$x=cos\theta-2cos^2\theta$
$y=sin\theta-2cos\theta sin\theta$

$\frac{dx}{d\theta}=-sin{\theta}(1-4cos{\theta})$

$\frac{dy}{d\theta}= cos -2cos2{\theta}$

I'm currently on the horizontal tangency but got stuck/wondering if there is another way:

$cos-2cos2{\theta}=0$

Thanks in advance

2. That looks good to me. $\frac{dy}{dx}= \frac{cos(\theta)- 2cos^2(\theta)}{-sin(\theta)(1- cos(\theta))}$.

The tangent will be horizontal when the numerator is 0:
[tex]cos(\theta)- 2cos^2(\theta)= cos(\theta)(1- 2cos(
\theta))= 0. That will be true wnen [tex]cos(\theta)= 0[tex] or when $cos(\theta)= \frac{1}{2}$.

The tangent will be vertical when the denominator is 0:
$-sin(\theta)(1- cos(\theta))= 0$ when $sin(\theta)= 0$ or $cos(\theta)= 1$.

Since both numerator and denominator are 0 when $cos(\theta)= 0$, that is not necessarily a vertical or horizontal tangent. You will need to look more closely at that.