Chapter 11.6, Problem 43E

a.

To determine

To show: that the polar equation of an ellipse with directrix $x=-d$ can be written in the form of $r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }$ .

Explanation of Solution

Given information:

The polar equation is $r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }$ .

Proof: since, the equation in standard form for directrix $x=-d$ is,

  $r=\frac{ed}{1-e\cos \theta }$

Since, it is known that $a^2=\frac{e^2{d}^2}{{\left(1-e^2\right)}^2}$ , so

  $\begin{array}{l}{a}^2=\frac{e^2{d}^2}{{\left(1-e^2\right)}^2}\\ a=\frac{ed}{\left(1-e^2\right)}\text{ [sqaure root]}\\ ed=a\left(1-e^2\right)\text{ [cross multiplication]}\end{array}$

Now, substitute in the standard form,

  $r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }$

b.

To determine

To find: an approximate polar equation for the elliptical orbit of the earth around the sun.

Answer to Problem 43E

The approximate polar equation for the elliptical orbit of the earth around the sun is $r=\frac{2.96\times {10}^8}{1-\left(0.017\right)\cos \theta }$ .

Explanation of Solution

Given information:

The eccentricity is $0.017$ and length of the major axis is about $2.99\times {10}^8\text{km}$ .

Calculation: to find the approximate polar equation, substitute in the formula,

  $\begin{array}{l}r=\frac{a\left(1-e^2\right)}{1-e\cos \theta }\\ r=\frac{\left(2.99\times {10}^8\right)\left(1-{\left(0.017\right)}^2\right)}{1-\left(0.017\right)\cos \theta }\text{ [substitute]}\\ r=\frac{2.96\times {10}^8}{1-\left(0.017\right)\cos \theta }\text{ [simplify]}\end{array}$

Thus, the polar equation is $r=\frac{2.96\times {10}^8}{1-\left(0.017\right)\cos \theta }$ .