Question

Question #48767

Answer 1

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)`

Answer 2

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.