Chapter 11.6, Problem 2E

To determine

To fill: The polar equations of a conic with eccentricity $e$ has one of the following forms:

$r=$________ or $r=$________.

Answer to Problem 2E

The polar equations of a conic with eccentricity $e$ has one of the following forms.

$r=\frac{ed}{1\pm e\cos \theta }$ or $r=\frac{ed}{1\pm e\sin \theta }$.

Explanation of Solution

Definition used:

Definition 1:

“The polar equation of a conic with focus at the origin and the directrix $x=d$ is $r=\frac{ed}{1+e\cos \theta }$”.

Definition 2:

“The polar equation of a conic with focus at the origin and the directrix $x=-d$ is $r=\frac{ed}{1-e\cos \theta }$”.

Definition 3:

“The polar equation of a conic with focus at the origin and the directrix $y=d$ is $r=\frac{ed}{1+e\sin \theta }$”.

Definition 4:

“The polar equation of a conic with focus at the origin and the directrix $y=-d$ is $r=\frac{ed}{1-e\sin \theta }$”.

Calculation:

From the definition 1, conclude that the polar equation of conic with directrix $x=d$ is,

$r=\frac{ed}{1+e\cos \theta }$ (1)

From the definition 2, conclude that the polar equation of conic with directrix $x=-d$ is

$r=\frac{ed}{1-e\cos \theta }$ (2)

From the equations (1) and (2) conclude that, the polar equation of conic with

directrix $x=\pm d$ is of the form $r=\frac{ed}{1\pm e\cos \theta }$.

From the definition 3, conclude that the polar equation of conic with directrix $x=d$ is

$r=\frac{ed}{1+e\sin \theta }$ (3)

From the definition 4, conclude that the polar equation of conic with directrix $y=-d$ is

$r=\frac{ed}{1-e\sin \theta }$ (4)

From the equations (3) and (4) conclude that, the polar equation of conic with

directrix $y=\pm d$ is of the form $r=\frac{ed}{1\pm e\sin \theta }$.

Thus, the polar equations of a conic with eccentricity $e$ has one of the following forms.

$r=\frac{ed}{1\pm e\cos \theta }$ or $r=\frac{ed}{1\pm e\sin \theta }$.