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.