Chapter 11, Problem 14T

(a)

To determine

To find: The polar equation of a conic and sketch the conic.

Answer to Problem 14T

The polar equation of the conic is $r=\frac{2}{2+\cos \theta }$.

Explanation of Solution

Given:

Polar region with focus at the origin, eccentricity as $\frac{1}{2}$ and directrix $x=2$.

Definition used:

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

Note 1:

“The conics $r=\frac{ed}{1\pm e\sin \theta }$ and $r=\frac{ed}{1\pm e\cos \theta }$ represents,

1. (i) An ellipse if $0<e<1$.
2. (ii) A parabola if $e=1$.
3. (iii) A hyperbola if $e>1$”.

Calculation:

The conic has the eccentricity $e=\frac{1}{2}$.

Thus, by the above note (1), the conic is an ellipse.

The given conic has its focus at the origin and directrix $x=2$.

By the above definition, the polar equation of conic is of the form $r=\frac{ed}{1+e\cos \theta }$.

Substitute $e=\frac{1}{2}$ and $d=2$ in $r=\frac{ed}{1+e\cos \theta }$.

$\begin{array}{c}r=\frac{ed}{1+e\cos \theta }\\ =\frac{\frac{1}{2}\left(2\right)}{1+\frac{1}{2}\cos \theta }\\ =\frac{2}{2+\cos \theta }\end{array}$

Therefore, the required equation of the conic is $r=\frac{2}{2+\cos \theta }$.

Obtain the graph the conic $r=\frac{2}{2+\cos \theta }$, obtain the intersection of the graph on the lines $\theta =0$, $\theta =\frac{\pi }{2}$, $\theta =\pi$ and $\theta =\frac{3\pi }{2}$ as follows.

Substitute $\theta =0$ in $r=\frac{2}{2+\cos \theta }$.

$\begin{array}{c}r=\frac{2}{2+\cos 0}\\ =\frac{2}{2+1}\\ =\frac{2}{3}\end{array}$.

Substitute $\theta =\frac{\pi }{2}$ in $r=\frac{2}{2+\cos \theta }$.

$\begin{array}{c}r=\frac{2}{2+\cos \theta }\\ =\frac{2}{2+\cos \frac{\pi }{2}}\\ =\frac{2}{2+0}\\ =1\end{array}$

Substitute $\theta =\pi$ in $r=\frac{2}{2+\cos \theta }$.

$\begin{array}{c}r=\frac{2}{2+\cos \theta }\\ =\frac{2}{2+\cos \pi }\\ =\frac{2}{2-1}\\ =2\end{array}$

Substitute $\theta =\frac{3\pi }{2}$ in $r=\frac{2}{2+\cos \theta }$

$\begin{array}{c}r=\frac{2}{2+\cos \theta }\\ =\frac{2}{2+\cos \frac{3\pi }{2}}\\ =\frac{2}{2+0}\\ =1\end{array}$

That is, the points of intersection on the lines $\theta =0$, $\theta =\frac{\pi }{2}$, $\theta =\pi$, $\theta =\frac{3\pi }{2}$ are shown below in the table.

$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r$	$\frac{2}{3}$	1	$2$	1

Use online graphing calculator to plot the points $\left(\frac{2}{3},0\right)$, $\left(1,\frac{\pi }{2}\right)$, $\left(2,\pi \right)$ and $\left(1,\frac{3\pi }{2}\right)$ on a polar graph and to the graph the equation $r=\frac{2}{2+\cos \theta }$ as shown below in Figure 1.

From Figure 1, it is observed that the obtained graph $r=\frac{2}{2+\cos \theta }$ is an ellispe.

(b)

To determine

To find: The type of conic represented by the polar equation $r=\frac{3}{2-\sin \theta }$.

Answer to Problem 14T

The type of conic is represented by the polar equation $r=\frac{3}{2-\sin \theta }$ is an ellipse.

Explanation of Solution

Definition used:

“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:

Given polar equation of conic is $r=\frac{3}{2-\sin \theta }$.

Rewrite the equation as follows,

$\begin{array}{c}r=\frac{3}{2-\sin \theta }\\ =\frac{\frac{3}{2}}{\frac{2}{2}-\frac{1}{2}\sin \theta }\left[\text{divide by 2}\right]\\ =\frac{\frac{3}{2}}{1-\frac{1}{2}\sin \theta }\end{array}$

Compare the equation $r=\frac{\frac{3}{2}}{1-\frac{1}{2}\sin \theta }$ with $r=\frac{ed}{1-e\sin \theta }$ then, it is observed that, $e=\frac{1}{2}$.

By the above note (1) equation $r=\frac{\frac{3}{2}}{1-\frac{1}{2}\sin \theta }$ is an ellipse.

Therefore, the conic $r=\frac{3}{2-\sin \theta }$ is an ellipse.

Obtain the graph of the equation $r=\frac{3}{2-\sin \theta }$, find the intercepts of the conic on the lines $\theta =0$, $\theta =\frac{\pi }{2}$, $\theta =\pi$ and $\theta =\frac{3\pi }{2}$ as follows.

Substitute $\theta =0$ in $r=\frac{3}{2-\sin \theta }$.

$\begin{array}{c}r=\frac{3}{2-\sin \theta }\\ =\frac{3}{2-\sin 0}\\ =\frac{3}{2-0}\\ =\frac{3}{2}\end{array}$

Substitute $\theta =0$ in $r=\frac{3}{2-\sin \theta }$.

$\begin{array}{c}r=\frac{3}{2-\sin \theta }\\ =\frac{3}{2-\sin \frac{\pi }{2}}\\ =\frac{3}{2-1}\\ =3\end{array}$

Substitute $\theta =\pi$ in $r=\frac{3}{2-\sin \theta }$.

$\begin{array}{c}r=\frac{3}{2-\sin \theta }\\ =\frac{3}{2-\sin \pi }\\ =\frac{3}{2-0}\\ =\frac{3}{2}\end{array}$

Substitute $\theta =\frac{3\pi }{2}$ in $r=\frac{3}{2-\sin \theta }$.

$\begin{array}{c}r=\frac{3}{2-\sin \theta }\\ =\frac{3}{2-\sin \frac{3\pi }{2}}\\ =\frac{3}{2+1}\\ =1\end{array}$.

That is, the points of intersection on the lines $\theta =0$, $\theta =\frac{\pi }{2}$ , $\theta =\pi$, $\theta =\frac{3\pi }{2}$ are shown below in table.

$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r$	$\frac{3}{2}$	3	$\frac{3}{2}$	$1$

Therefore, polar coordinates are $\left(\frac{3}{2},0\right)$, $\left(3,\frac{\pi }{2}\right)$, $\left(\frac{3}{2},\pi \right)$, $\left(1,\frac{3\pi }{2}\right)$.

Graph:

Use online graphing calculator to draw the equation of $r=\frac{3}{2-\sin \theta }$ by plotting the above coordinates shown below in Figure 2.

From Figure 2, it is observed that the required Thus, the sketch of the conic $r=\frac{3}{2-\sin \theta }$ is the conic is an ellipse.