Chapter 9.4, Problem 32PPS

To determine

To write: an equation relating the comet’s orbit around the sun in millions of miles with the centre of the horizontal ellipse at the origin.

Answer to Problem 32PPS

The equation for the given ellipse is $\frac{x^2}{{\left(1672.82\right)}^2}+\frac{y^2}{{\left(426.58\right)}^2}=1$ .

Explanation of Solution

Given information:

The given aphelion of the elliptical orbit of the sun is $3282.9\text{millions}\text{miles}$ and perihelion is $54.87\text{millions}\text{miles}$ .

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 planet Halley’s Comet travels around the sun in elliptical orbit.

The aphelion of the elliptical orbit of the sun is $3282.9\text{millions}\text{miles}$ and perihelion is $54.87\text{millions}\text{miles}$

The diameter of the sun is $0.87\text{millions}\text{miles}$

The length of major axis of the ellipse is the sum of aphelion, perihelion and sun’s diameter.

Simplify as,

  $\begin{array}{l}2a=3282.9+54.87+0.87\\ 2a=3345.64\\ a=1672.82\text{millions}\text{miles}\end{array}$

The focus of the centre is the value of $a=1672.82$ minus the sum of radius of the sun and perihelion

  $\begin{array}{l}c=1672.82-\left(\frac{0.87}{2}+54.87\right)\\ c=1672.82-55.305\\ c=1617.515\end{array}$

To calculate the value of $b$ by using the relations between of the $a,b$ and $c$ is

  $a^2-b^2=c^2$

  $\begin{array}{l}{\left(1672.82\right)}^2-b^2={\left(1617.515\right)}^2\\ 2798326.7524-b^2=2616354.775225\\ b^2=181971.977175\\ b=426.58\end{array}$

The standard form of the equation at the given origin centre of the elliptical orbit of the planet Halley’s comet is,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(1672.82\right)}^2}+\frac{y^2}{{\left(426.58\right)}^2}=1\end{array}$

Thus, the equation for the given ellipse is $\frac{x^2}{{\left(1672.82\right)}^2}+\frac{y^2}{{\left(426.58\right)}^2}=1$ .