# Question #48767

Feb 13, 2018

Thus,
$k = \pm \left(r - 1\right)$

#### Explanation:

$r = 1 - k \cos t$
Here, $\cos t$ can change from -1 to 1
For $\cos t = - 1$
$r = 1 - k \left(- 1\right)$
$r = 1 + k$
Solving for k
$k = r - 1$

for $\cos t = 1$
$r = 1 - k \left(1\right)$
$r = 1 - k$
Solving for k
$k = 1 - r$

Thus,
$k = \pm \left(r - 1\right)$

Feb 13, 2018

The graph is that of
$r \left(t\right) = 1 - 3 \cos \left(t\right)$, so $k = 3$

#### Explanation:

Let us take a look at the points at which the curve cuts the $X$ axis for nonzero $r$. These are the points with Cartesian coordinates $\left(- 2 , 0\right)$ and $\left(- 4 , 0\right)$, respectively.

One of them correspond to $t = 0$, the other to $t = \pi$. The $r$ values for these two points must be $1 - k$ and $1 + k$, respectively. Of these, the first must be negative (a positive $r$ for $t = 0$ would lead to a point to the right of the origin), leading to a distance from the origin of $k - 1$. Since this is smaller than $k + 1$, this must correspond to
$\left(- 2 , 0\right)$

(The above follows simply from the correspondence $x = r \cos \left(t\right) , y = r \sin \left(t\right)$ between polar and Cartesian coordinates.)

Thus

$- 2 = 1 - k \cos \left(0\right) = 1 - k$

This will lead to $k = 3$

A check : note that this is consistent with $r \left(\pi\right) = 4$ - the other point on the $X$ axis.