Question #48767

2 Answers
Feb 13, 2018

Thus,
k=+-(r-1)

Explanation:

r=1-kcost
Here, cost can change from -1 to 1
For cost=-1
r=1-k(-1)
r=1+k
Solving for k
k=r-1

for cost=1
r=1-k(1)
r=1-k
Solving for k
k=1-r

Thus,
k=+-(r-1)

Feb 13, 2018

The graph is that of
r(t)=1-3cos(t), 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 (-2,0) and (-4,0), 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
(-2,0)

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

Thus

-2=1-k cos(0) = 1-k

This will lead to k =3

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