Chapter 8, Problem 58RE

Solution

ToIdentify:The conic $r=\frac{3}{4+\sin \theta }$ , graph it and rewrite the equation in Cartesian co-ordinates.

The conic $r=\frac{3}{4+\sin \theta }$ is an Ellipse.

The graph of the conic is:

  

The equation of the conic in Cartesian co-ordinates is: $\frac{25{\left(y+\frac{1}{5}\right)}^2}{16}+\frac{5x^2}{3}=1$ .

Given information:

The polar equation of the conic is: $r=\frac{3}{4+\sin \theta }$ .

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 $0\le e<1$ , the conic is an Ellipse.

The standard equation of an Ellipse:

• Having $y=k$ as its Focal axis (the larger number is the denominator of $y^2$):

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Here, $\left(h,k\right)$ are the coordinates of the center of the Ellipse.

  $\left(h\pm a,k\right)$ are the coordinates of the vertices of the Ellipse.

  $a$ is the semi-major axis and $b$ is the semi minor axis,

  $\left(h\pm c,k\right)$ are the coordinates of the Foci of the Ellipse.

The Pythagorean relation of Ellipse is $a^2=b^2+c^2$ .

• Having $x=h$ as its Focal axis (the larger number is the denominator of $x^2$):

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

Here, $\left(h,k\right)$ are the coordinates of the center of the Ellipse.

  $\left(h,k\pm a\right)$ are the coordinates of the vertices of the Ellipse.

  $a$ is the semi-major axis and $b$ is the semi minor axis,

  $\left(h,k\pm c\right)$ are the coordinates of the Foci of the Ellipse.

The Pythagorean relation of Ellipse is $a^2=b^2+c^2$ .

Calculation:

1. To determine the type of conic:

The polar equation of the conic is $r=\frac{3}{4+\sin \theta }$ .

Divide the numerator and the denominator by $4$ to get the equation in standard form of the polar equation of conics:

  $\begin{array}{l}r=\frac{\frac{3}{4}}{\frac{4+\sin \theta }{4}}\\ r=\frac{\frac{3}{4}}{1+\frac{1}{4}\sin \theta }\\ r=\frac{\frac{1}{4}.3}{1+\frac{1}{4}\sin \theta }\end{array}$

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

  $e=\frac{1}{4}$ and $p=3$ .

Since, $0\le e<1$ , the conic is an ellipse with directrix $x=p=3$ .

Thus, the conic $r=\frac{3}{4+\sin \theta }$ represents an Ellipse.

2. To graph the conic $r=\frac{3}{4+\sin \theta }$:

The polar equation of the conic is $r=\frac{3}{4+\sin \theta }$ .

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:

$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r=\frac{3}{4+\sin \theta }$	$\frac{3}{4}$	$\frac{3}{5}$	$-1$	$1$

Plot the points to create a graph of the Ellipse:

  

Figure(1)

3. Rewrite the equation of the conic in Cartesian coordinates:

The conic $r=\frac{3}{4+\sin \theta }$ represents an Ellipse with eccentricity, $e=\frac{1}{4}$ .

Let $2a$ be the length of the major axis and $2b$ be the length of the minor axis of the ellipse $r=\frac{3}{4+\sin \theta }$ .

According to Figure(1),

The end points of the major axis are $\left(0,\frac{3}{5}\right)=\left(0,0.06\right)$ and $\left(0,-1\right)$ .

The length of the major axis, $2a$ is given by:

  $\begin{array}{l}2a=\left|\frac{3}{5}-\left(-1\right)\right|\\ 2a=\left|\frac{3}{5}+1\right|\\ 2a=\frac{8}{5}\\ a=\frac{4}{5}\end{array}$

Since the center lies in the midpoint of the two vertices of the Ellipse viz. midpoint of the major axis

The coordinates of the center $\left(h,k\right)$ are:

  $\begin{array}{l}\left(h,k\right)=\left(\frac{0+0}{2},\frac{\frac{3}{5}+\left(-1\right)}{2}\right)\\ \left(h,k\right)=\left(0,\frac{\frac{3}{5}-1}{2}\right)\\ \left(h,k\right)=\left(0,\frac{\frac{-2}{5}}{2}\right)\\ \left(h,k\right)=\left(0,-\frac{1}{5}\right)\end{array}$

The eccentricity, $e$ of the Ellipse is given by $e=\frac{c}{a}$ .

Substitute $e=\frac{1}{4}$ and $a=\frac{4}{5}$ in the formula of Eccentricity to obtain the value of $c$ :

  $\begin{array}{l}\frac{1}{4}=\frac{c}{\frac{4}{5}}\\ \frac{1}{4}=\frac{5c}{4}\\ 1=5c\\ c=\frac{1}{5}\end{array}$

Substitute $a=\frac{4}{5}$ and $c=\frac{1}{5}$ in the Pythagorean Identity, $a^2=b^2+c^2$ of the Ellipse to obtain the value of $b^2$ :

  $\begin{array}{l}{\left(\frac{4}{5}\right)}^2=b^2+{\left(\frac{1}{5}\right)}^2\\ \frac{16}{25}=b^2+\frac{1}{25}\\ b^2=\frac{16}{25}-\frac{1}{25}\\ b^2=\frac{15}{25}\end{array}$

Simplify further to obtain:

  $b^2=\frac{3}{5}$

Substitute $\left(h,k\right)=\left(0,-\frac{1}{5}\right)$ , $a=\frac{4}{5}$ and $a=\frac{4}{5}$ in the formula for the Cartesian equation of ellipse:

  $\begin{array}{l}\frac{{\left(y-\left(-\frac{1}{5}\right)\right)}^2}{{\left(\frac{4}{5}\right)}^2}+\frac{{\left(x-0\right)}^2}{\frac{3}{5}}=1\\ \frac{{\left(y+\frac{1}{5}\right)}^2}{\frac{16}{25}}+\frac{x^2}{\frac{3}{5}}=1\\ \frac{25{\left(y+\frac{1}{5}\right)}^2}{16}+\frac{5x^2}{3}=1\end{array}$

Thus, the equation of the conic $r=\frac{3}{4+\sin \theta }$ in Cartesian coordinates is $\frac{25{\left(y+\frac{1}{5}\right)}^2}{16}+\frac{5x^2}{3}=1$ .