Chapter 8, Problem 62RE

Solution

ToIdentify:The conic $r=\frac{4}{4-4\cos \theta }$ , graph it and rewrite the equation in Cartesian co-ordinates.

The conic $r=\frac{4}{4-4\cos \theta }$ is a parabola.

The graph of the conic is:

  

The equation of the conic in Cartesian co-ordinates is: $y^2=2\left(x+\frac{1}{2}\right)$

Given information:

The polar equation of the conic is: $r=\frac{4}{4-4\cos \theta }$ .

Formula used:

Polar equation of conics:

For a conic with focus at the origin, if the directrix is  $x=\pm p$ , where  $p$  is a positive real number, and the 

eccentricity is a positive real number  $e$ ,

Then, the polar equation of the conic is:

  $r=\frac{ep}{1\pm e\cos \theta }$

For a conic with focus at the origin, if the directrix is  $y=\pm p$ , where  $p$  is a positive real number, and the 

eccentricity is a positive real number  $e$ ,

Then, the polar equation of the conic is:

  $r=\frac{ep}{1\pm e\sin \theta }$

Determination of the conic according to the value of the eccentricity, $e$ of the conic:

If $e>1$ , the conic is a Hyperbola.

If $e=1$ , the conic is a Parabola.

If $e<1$ , the conic is an Ellipse.

The standard equation of a horizontal parabola:

  ${\left(y-k\right)}^2=4p\left(x-h\right)$

Here, $\left(h,k\right)$ are the coordinates of the vertex of the parabola,

  $p$ is the Focal length of the parabola,

  $\left(h+p,k\right)$ are the coordinates of the Focus of the parabola.

Calculation:

1. To determine the type of conic:

The polar equation of the conic is $r=\frac{4}{4-4\cos \theta }$ .

Write the equation in standard form of the polar equation of conics:

  $\begin{array}{l}r=\frac{4}{4-4\cos \theta }\\ r=\frac{4}{4\left(1-\cos \theta \right)}\\ r=\frac{1}{1-\cos \theta }\\ r=\frac{1\cdot 1}{1-1\cdot \cos \theta }\end{array}$

Compare the above equation with the standard polar equation of conics $r=\frac{ep}{1\pm e\cos \theta }$ to obtain:

  $e=1$ and $p=1$ .

Since, $e=1$ , the conic is a Parabola with directrix $x=p$ i.e. $x=1$ .

Thus, the conic $r=\frac{4}{4-4\cos \theta }$ represents a Parabola.

2. To graph the conic $r=\frac{4}{4-4\cos \theta }$:

The polar equation of the conic is $r=\frac{4}{4-4\cos \theta }$ .

Create a table to obtain the vertices by substituting the values of $0\le \theta \le 2\pi$ to get the corresponding values of r:

$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r=\frac{4}{4-4\cos \theta }$	undefined	$1$	$\frac{1}{2}=0.5$	$\frac{1}{2}=0.5$

Plot the points to create a graph of the parabola :

  

Figure(1)

According to Figure(1),

The vertex of the parabola is at $\left(-\frac{1}{2},0\right)$ and the focus is at the origin $O\left(0,0\right)$ .

3. Rewrite the equation of the conic $r=\frac{4}{4-4\cos \theta }$ in Cartesian coordinates:

The conic $r=\frac{4}{4-4\cos \theta }$ represents a parabola with the vertex at $\left(-\frac{1}{2},0\right)$ and focus at the origin $O\left(0,0\right)$ .

Substitute the coordinates of the vertex $\left(h,k\right)=\left(-\frac{1}{2},0\right)$ and the coordinates of the focus as $\left(0,0\right)$ in the Standard equation of a Horizontal parabola, ${\left(y-k\right)}^2=4p\left(x-h\right)$ to obtain its equation in Cartesian coordinates:

  $\begin{array}{l}{\left(y-0\right)}^2=4p\left(x-\left(-\frac{1}{2}\right)\right)\\ y^2=4p\left(x+\frac{1}{2}\right)\end{array}$

The coordinates of the Focus of the parabola are $\left(h+p,k\right)$ i.e. $\left(h+p,k\right)=\left(0,0\right)$ .

Substitute $h=-\frac{1}{2}$ in $h+p=0$ to obtain the value of $p$ :

  $\begin{array}{l}h+p=0\\ -\frac{1}{2}+p=0\\ p=\frac{1}{2}\end{array}$

Substitute the value $p=\frac{1}{2}$ in the obtained equation $y^2=4p\left(x+\frac{1}{2}\right)$ of the parabola in Cartesian coordinates:

  $\begin{array}{l}{y}^2=4p\left(x+\frac{1}{2}\right)\\ y^2=4\left(\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\\ y^2=2\left(x+\frac{1}{2}\right)\end{array}$

Thus, the equation of the conic $r=\frac{4}{4-4\cos \theta }$ in Cartesian coordinates is $y^2=2\left(x+\frac{1}{2}\right)$ .