Chapter 9.4, Problem 5E

To determine

To find: The ratio of the length of the major axis of this orbit to the length of the minor axis.

Answer to Problem 5E

Therefore, the foci of ellipse is $\left(0,\sqrt{6}\right),\left(0,-\sqrt{6}\right)$ .

Explanation of Solution

Given information:

The mean distance of the asteroid from the sun is $1.078\text{AU}$ and the eccentricity of the orbit is $0.827$ .

The given equation is,

  $3x^2+y^2=\mathrm{1.....}(1)$ .

The equation of ellipse is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Divide equation (1) by $9$ .

  $\frac{x^2}{3}+\frac{y^2}{9}=\mathrm{1....}(2)$ .

It can be rewritten as,

  $\frac{x^2}{{\left(\sqrt{3}\right)}^2}+\frac{y^2}{3^2}=1$ .

Now it is similar t the equation of ellipse. On comparing,

  $\begin{array}{l}a=3\\ b=\sqrt{3}\end{array}$

Since the denominator of $y^2$ is larger so the major axis is vertical.

Now, determine the x - intercept and y -intercept by replacing the $y$ with $0$ and $x$ with $0$ respectively.

The x -intercept is,

  $\begin{array}{c}\frac{x^2}{\sqrt{3^2}}+\frac{0}{3^2}=1\\ x^2={\left(\sqrt{3}\right)}^2\\ x=\sqrt{3}\end{array}$ .

The y -intercept is,

  $\begin{array}{c}\frac{0}{\sqrt{3^2}}+\frac{y^2}{3^2}=1\\ y^2=3^2\\ y=3\end{array}$

Equation (2) can be written as,

  $\begin{array}{c}\frac{x^2}{3}+\frac{y^2}{9}=1\\ \frac{y^2}{9}=1-\frac{x^2}{3}\\ y^2=\frac{9}{3}\left(3-x^2\right)\\ y=\pm \sqrt{3\left(3-x^2\right)}\end{array}$

It can be seen that $\left|x\right|$ is a real number if and only if $\left|y\right|\le 3$ .

Thus, The extent of the graph is limited to x - intercept $\sqrt{3}$ and $\sqrt{-3}$ , and the y -intercept is $3$ and $-3$ .

Now make the table of the points in the first quadrant as the graph is symmetric being the centre of the ellipse is $\left(0,0\right)$ .

  

The graph of ellipse is,

  

Use the relationship,

  $c^2=a^2-b^2$ .

Here substitute $a=3,b=\sqrt{3}$ .

  $\begin{array}{c}{c}^2=a^2-b^2\\ =3^2-{\left(\sqrt{3}\right)}^2\\ 9-3\\ =6\end{array}$

On taking square root on both sides,

  $\begin{array}{c}{c}^2=\sqrt{6}\\ c=\sqrt{6}\end{array}$

Therefore, the foci of ellipse is $\left(0,\sqrt{6}\right),\left(0,-\sqrt{6}\right)$ .