Chapter 8.5, Problem 10E

Solution

To find:The eccentricity, type of conic and the directrix.

Theconic is an ellipse, with eccentricity $e=\frac{1}{4}$ and directrix $x=-2$ .

Given information:

The given polar equation is $r=\frac{2}{4-\cos \theta }$ .

Formula used:

  $e<1$ , the conic is an ellipse.

  $e=1$ , the conic is a parabola.

  $e>1$ , the conic is a hyperbola.

  $r=\frac{ke}{1+e\cos \theta }$ when directrix is vertical, to the right of the pole: $x=k$ .

  $r=\frac{ke}{1-e\cos \theta }$ when directrix is vertical, to the right of the pole: $x=-k$ .

  $r=\frac{ke}{1+e\sin \theta }$ when directrix is horizontal, to the right of the pole: $y=k$ .

  $r=\frac{ke}{1-e\sin \theta }$ when directrix is horizontal, to the right of the pole: $y=-k$ .

Calculation:

Compare $r=\frac{2}{4-\cos \theta }$ with $r=\frac{ke}{1-e\cos \theta }$ .

Take $4$ common in the denominator.

  $r=\frac{\frac{2}{4}}{1-\frac{1}{4}\cos \theta }$

Here, $k=2$ and $e=\frac{1}{4}$ .

Since, $e<1$ , the conic is an ellipse.

The equation of directrix is $x=-k$ .

Substitute $2$ for $k$ in $x=-k$ .

Thus, equation of directrix for $r=\frac{2}{4-\cos \theta }$ is $x=-2$ .

Therefore, the conic is an ellipse, with eccentricity $e=\frac{1}{4}$ and directrix $x=-2$ .