Chapter 9.4, Problem 24WE

(a)

To determine

To explain: e a number between 0 and 1.

Answer to Problem 24WE

  $0<\pm e<1$ Proved that e is a number between 0 and 1.

Explanation of Solution

Given:

Suppose that $a=b$ in the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Calculation:

The given equation of the ellipse is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The eccentricity is e where $e=\frac{c}{a}$ and $c=\sqrt{a^2-b^2}$ .

Explain why e is a number between 0 and 1.

From the given equation of the ellipse $a>b$ and a and b are real numbers. So a, b>0.

Hence write as

  $0<b<a$

From this write as $a^2-b^2>0$

And

  $a^2-b^2<a^2\left[\text{Since we are subtracting a real number }{b}^2\text{ from }{a}^2\right]\mathrm{......}\left(1\right)$

Using equation (1),

  $\begin{array}{l}0<a^2-b^2<a^2\hfill \\ 0<c^2<a^2\left[\text{since }c=\sqrt{a^2-b^2}\right]\hfill \end{array}$

Dividing all the terms by $a^2$ ,

  $\begin{array}{l}0<\frac{c^2}{a^2}<\frac{a^2}{a^2}\\ 0<\frac{c^2}{a^2}<1\end{array}$

By taking square root in all the terms,

  $\begin{array}{l}0<\pm \frac{c}{a}<1\\ 0<\pm e<1\left[\text{since }e=\frac{c}{a}\right]\end{array}$

The e value cannot be negative since e is between 0 and 1.

Therefore, $0<\pm e<1$

Thus proved that e is a number between 0 and 1.

Conclusion:

Therefore, $0<\pm e<1$ proved that e is a number between 0 and 1.

(b)

To determine

To find: the ellipses shown above has the greater eccentricity.

Answer to Problem 24WE

Therefore the third figure has greater eccentricity

Explanation of Solution

Calculation:

Find that which has greater eccentricity

  

  

  

In the first figure, the length of the major axis is equal to the length of the minor axis.

Therefore, $a=b$ .

Now

  $c=\sqrt{a^2-b^2}$

By squaring in both sides,

  $\begin{array}{l}{c}^2=a^2-b^2\hfill \\ =a^2-a^2\left[Sincea=b\right]\hfill \\ =0\hfill \end{array}$

Therefore, $c=0$

Again, $e=\frac{c}{a}$

Since $c=0$ ,

Thus, $e=0$

In the second figure, the length of the major axis greater than the length of the minor axis.

Therefore, $a>b$ .

Now

  $c=\sqrt{a^2-b^2}$

By squaring in both sides,

  $\begin{array}{l}{c}^2=a^2-b^2\\ a^2=c^2+b^2\end{array}$

Dividing both sides by $a^2$ ,

  $\begin{array}{l}\frac{a^2}{a^2}=\frac{c^2}{a^2}+\frac{b^2}{a^2}\\ 1=e^2+\frac{b^2}{a^2}\left[\text{since }e=\frac{c}{a}\right]\\ e=\sqrt{1-\frac{b^2}{a^2}}\left[e\text{ cannot be negative}\right]\end{array}$

Since $a>b$ , $1-\frac{b^2}{a^2}$ is a number less than 1.

Therefore, $e<1$ .

In the third figure, the length of the major axis is little less than the length of the minor axis. In this case both the axis is almost equal.

Therefore, $a\approx b$

Now

  $c=\sqrt{a^2-b^2}$

By squaring in both sides,

  $\begin{array}{l}{c}^2=a^2-b^2\\ a^2=c^2+b^2\end{array}$

Dividing both sides by $a^2$ ,

  $\begin{array}{l}\frac{a^2}{a^2}=\frac{c^2}{a^2}+\frac{b^2}{a^2}\\ 1=e^2+\frac{b^2}{a^2}\left[\text{since }e=\frac{c}{a}\right]\\ e=\sqrt{1-\frac{b^2}{a^2}}\left[e\text{ cannot be negative}\right]\end{array}$

Since $a\approx b$ , $1-\frac{b^2}{a^2}$ term is almost equal to 1.

Therefore, e=1.

Conclusion:

Therefore the third figure has greater eccentricity.