Chapter 10.3, Problem 76E

(a)

To determine

To find: Graph the function in a square viewing window. Describe what you see.

Answer to Problem 76E

It is observed in the graph that the eccentricity $e=1.2$ .The asymptote of the hyperbola are $\pm {\cos }^{-1}\left(\frac{-1}{1.2}\right)$ .

Explanation of Solution

Given information:

The polar equation is $r=\frac{1}{1+1.2\cos \theta }$

Calculation:

The given equation is polar equation of ellipse. The standard polar equation of ellipse is $r=\frac{c}{1+e\cos \theta }$

The viewing window will be in ratio $3:2$ .Let the viewing window be $\left[-5,5\right],\left[4.5,4.5\right]$

The graph of the given function in square window is below:

  

It is observed in the graph that the eccentricity $e=1.2$ therefore the ellipse becomes a hyperbola with two branches. The asymptote of the hyperbola is $\pm {\cos }^{-1}\left(\frac{-1}{1.2}\right)$ .

The vertex of the hyperbola is at $\left(0.45,0),(5,0)\right)$ both are at positive side of $x$ axis.

(b)

To determine

To explain: why do you get both branches of parabola.

Answer to Problem 76E

Explanation is given.

Explanation of Solution

Given information:

The polar equation of ellipse is $r=\frac{1}{1+1.2\cos \theta }$ .

Calculation:

The given equation is polar equation of ellipse. The standard polar equation of ellipse is $r=\frac{c}{1+e\cos \theta }$

Here the constant $e$ is the eccentricity. It is the shape of the orbit.

Compare the given equation with the standard equation will give $e=1.2$ .

If $e=1$ then the curve opens up to a parabola and if $e>1$ the parabola opens up to branch of the hyperbola with the distance approaching to infinity as $1+e\cos \theta$ approaches to zero.

Therefore, the it gives two branches of parabola because it becomes hyperbola as eccentricity is greater than one.

(c)

To determine

To find: why the graphing calculator appears to show the asymptote.

Answer to Problem 76E

Explanation is given.

Explanation of Solution

Given information:

The polar equation of ellipse is $r=\frac{1}{1+1.2\cos \theta }$

Calculation:

The given equation is polar equation of ellipse. The standard polar equation of ellipse is $r=\frac{c}{1+e\cos \theta }$ .

The asymptotes of the curves are $\theta =\pm {\cos }^{-1}\left(\frac{-1}{1.2}\right)$ .

  $\begin{array}{l}\theta =\pm {\cos }^{-1}\left(\frac{-1}{1.2}\right)\\ \theta =\pm 146.4\end{array}$

In polar curves $r$ is the distance origin to the curve and $\theta$ is the angle from the initial ray as of positive direction of $x$ axis.

All the asymptotes lie between $0to2\pi$ Therefore graphing window show all the asymptote because in polar coordinate $x,y$ axis represents the angle $\theta$ .