Chapter 10.6, Problem 11AYU

To determine

The conic that the polar equation represents and give the position of directrix.

Answer to Problem 11AYU

ellipse,

Directrix is perpendicular to the polar axis at distance $3/2$ units to the left of the pole.

Explanation of Solution

Given:

$r=\frac{3}{4-2\text{cos}\theta }$

Formula used:

Polar Equation	Eccentricity	Directrix
$r=\frac{ep}{1-e\text{cos}\theta }$	$e=1:\text{ Parabola};e<1:\text{ ellipse};e>1:\text{ hyperbola}$	Directrix is perpendicular to the polar axis at distance $p$ units to the left of the pole.
$r=\frac{ep}{1\text{ + }e\text{cos}\theta }$	$e=1:\text{ Parabola};e<1:\text{ ellipse};e>1:\text{ hyperbola}$	Directrix is perpendicular to the polar axis at distance $p$ units to the right of the pole.
$r=\frac{ep}{1\text{ + }e\text{sin}\theta }$	$e=1:\text{ Parabola};e<1:\text{ ellipse};e>1:\text{ hyperbola}$	Directrix is parallel to the polar axis at distance $p$ units above the pole.
$r=\frac{ep}{1-e\text{sin}\theta }$	$e=1:\text{ Parabola};e<1:\text{ ellipse};e>1:\text{ hyperbola}$	Directrix is parallel to the polar axis at distance $p$ units below the pole.

Calculation:

$r=\frac{3}{4-2\text{cos}\theta }$ rewriting the equation,

$r=\frac{\frac{3}{4}}{1-\frac{1}{2}\text{cos}\theta }$ is in the form of $r=\frac{ep}{1+e\text{cos}\theta }$.

Hence, $ep=\frac{3}{4};e=\frac{1}{2}$ and $p=3/2$.

Since $e=\frac{1}{2}<1$,

Therefore, $r=\frac{3}{4-2\text{cos}\theta }$ is an ellipse.

Directrix is perpendicular to the polar axis at distance $3/2$ units to the left of the pole.