Chapter 8.4, Problem 51PS

Solution

To calculate : The values of $a\text{ and }c$ and an equation for orbit.

The required equation is $\frac{x^2}{{\left(17.94\right)}^2}+\frac{y^2}{{\left(20.82\right)}^2}=1$ and the values are $a=17.94\text{ and }c=17.35$ .

Given information :

Halley's comet ranges from $0.59\text{ to }35.3$ astronomical units from the sun, which is at focus of the comet's elliptical orbit.

Calculation :

Draw the figure on the basis of the given information.

  

The sum of the distances from a point on the ellipse to the foci is a constant. When the comet is at a distance of $0.59$ units, it must be at the vertex near to the sun and when it is at a distance of $35.3$ units, it must be at the vertex farthest from the sun.

The total distance between two vertices is denoted by $2a$ . So,

  $\begin{array}{c}2a=0.59+35.3\\ =35.89\\ a=17.94\end{array}$

Each focus is units away from the center. The distance of a vertex from the center is $17.94$ units. Since one focus is $0.59$ units away from the vertex, so

  $\begin{array}{c}c=17.94-0.59\\ =17.35\end{array}$

The equation of an ellipse is in the form of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . Since, the major axis is horizontal, so

  $\begin{array}{c}{c}^2=a^2-b^2\\ {\left(17.35\right)}^2={\left(17.94\right)}^2-b^2\\ b^2=20.82\end{array}$

So, the equation is $\frac{x^2}{{\left(17.94\right)}^2}+\frac{y^2}{{\left(20.82\right)}^2}=1$ .

Therefore, the required equation is $\frac{x^2}{{\left(17.94\right)}^2}+\frac{y^2}{{\left(20.82\right)}^2}=1$ and the values are $a=17.94\text{ and }c=17.35$ .