Chapter 8.3, Problem 55E

Solution

The value of $a$ , $b$ , $c$ , $e$ and the coordinates of the center of the Sun if the center of the hyperbolic orbit is $\left(0,0\right)$ and the Sun lies on the positive $x$ axis.

It has been determined that $a=1440$ , $b=600$ , $c=1560$ and $e\approx 1.083$ .

It has also been determined that the coordinates of the center of the Sun are $\left(1560,0\right)$ .

Given:

A comet following a hyperbolic path about the Sun has a perihelion of $120$ Gm. When the line from the comet to the Sun is perpendicular to the focal axis of the orbit, the comet is $250$ Gm from the Sun.

Concept used:

The perihelion distance of a hyperbola is $c-a$ .

Calculation:

It is given that a comet follows a hyperbolic path about the Sun such that the center of the hyperbolic orbit is $\left(0,0\right)$ and the Sun lies on the positive $x$ axis.

Then, the equation of the hyperbola is given as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

As discussed, the perihelion distance of a hyperbola is $c-a$ .

It is given that the comet following a hyperbolic path about the Sun has a perihelion of $120$ Gm.

Then, $c-a=120$ .

Now, for a hyperbola, $c^2=a^2+b^2$ .

Put $x=c$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get,

  $\begin{array}{c}\frac{c^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{y^2}{b^2}=\frac{c^2}{a^2}-1\\ y^2=b^2\left(\frac{c^2}{a^2}-1\right)\end{array}$

Put $c^2=a^2+b^2$ in $y^2=b^2\left(\frac{c^2}{a^2}-1\right)$ to get,

  $\begin{array}{c}{y}^2=b^2\left(\frac{a^2+b^2}{a^2}-1\right)\\ =b^2\left(1+\frac{b^2}{a^2}-1\right)\\ =b^2\left(\frac{b^2}{a^2}\right)\\ =\frac{b^4}{a^2}\end{array}$

So, $y=\pm \frac{b^2}{a}$ .

It is given that when the line from the comet to the Sun is perpendicular to the focal axis of the orbit, the comet is $250$ Gm from the Sun.

This implies that $\frac{b^2}{a}=250$ .

Simplifying,

  $b^2=250a$

Put $b^2=250a$ in $c^2=a^2+b^2$ to get,

  $c^2=a^2+250a$

Simplifying,

  $c^2-a^2=250a$

On further simplification,

  $\left(c-a\right)\left(c+a\right)=250a$

Put $c-a=120$ in the above equation to get,

  $120\left(c+a\right)=250a$

Simplifying,

  $120c+120a=250a$

On further simplification,

  $120c=130a$

Now, $c-a=120$ . Then, $c=120+a$ .

Put $c=120+a$ in $120c=130a$ to get,

  $\begin{array}{c}120\left(120+a\right)=130a\\ 14400+120a=130a\\ 10a=14400\\ a=1440\end{array}$

So, $a=1440$ .

Put $a=1440$ in $b^2=250a$ to get,

  $b^2=250\left(1440\right)$

Simplifying,

  $b=\pm 600$

Taking the positive value, $b=600$ .

Put $a=1440$ in $c=120+a$ to get,

  $c=1560$

Put $c=1560$ and $a=1440$ in $e=\frac{c}{a}$ to get,

  $e=\frac{1560}{1440}$

Simplifying,

  $e\approx 1.083$

Now, the Sun is at the focus, which is the point $\left(c,0\right)$ .

Then, the coordinates of the center of the Sun are $\left(1560,0\right)$ .

Conclusion:

It has been determined that $a=1440$ , $b=600$ , $c=1560$ and $e\approx 1.083$ .

It has also been determined that the coordinates of the center of the Sun are $\left(1560,0\right)$ .