Chapter 9.4, Problem 3GP

To determine

To find: The equation relating Pluto’s orbit around the Sun in millions of miles.

Answer to Problem 3GP

The equation relating Pluto’s orbit around the Sun in millions of miles is $\frac{x^2}{{\left(4.583\text{billion}\right)}^2}+\frac{y^2}{{\left(4.203\text{billion}\right)}^2}=1$

Explanation of Solution

Given information:

Pluto’s orbit is about $2.757\text{billon}$ miles at perihelion and about $4.583\text{millon}$ miles at aphelion.

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

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

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

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

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The diameter of the Sun is $0.87\text{million}$ miles.

The value of $a$ is calculated as,

  $\begin{array}{l}2a=0.00087+2.757+4.583\\ 2a=7.34087\\ a\approx 3.670\text{billion miles}\end{array}$

The value of $c$ is calculated as,

  $\begin{array}{c}c=4.583-\left(2.757+\frac{0.00087}{2}\right)\\ =1.826\text{billion miles}\end{array}$

The value of $b$ is calculated as,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ 1.826=\sqrt{{\left(4.583\right)}^2-b^2}\\ b=4.203\text{billion miles}\end{array}$

The standard form of the equation of the given ellipse is,

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-0\right)}^2}{{\left(4.583\text{billion}\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(4.203\text{billion}\right)}^2}=1\\ \frac{x^2}{{\left(4.583\text{billion}\right)}^2}+\frac{y^2}{{\left(4.203\text{billion}\right)}^2}=1\end{array}$

Thus, the equation for the given ellipse is $\frac{x^2}{{\left(4.583\text{billion}\right)}^2}+\frac{y^2}{{\left(4.203\text{billion}\right)}^2}=1$ .