Chapter 9.4, Problem 4E

To determine

To find: The location of the planet is from the sun in AU.

Explanation of Solution

Given information:

Mercury is on the y-axis in the coordinate system.

The given equation of the ellipse is,

  $x^2+4y^2=\mathrm{16.....}(1)$ .

The equation of ellipse is,

  $\frac{x^2}{a^2}+\frac{y2}{b^2}=1$ .

Divide both side of equation (1) by $16$ ,

  $\frac{x^2}{16}+\frac{4y2}{16}=\mathrm{1....}(2)$ .

It can be written as,

  $\frac{x^2}{4^2}+{\frac{y}{2^2}}^2=1$ .

Since the denominator of $x^2$ term is larger so the major axis is horizontal and minor axis is vertical.

We have the relation,

  $c^2=a^2-b^2\mathrm{.....}(3)$ .

Here,

  $\begin{array}{l}{c}^2=12\\ b^2=4\end{array}$

Substitute in equation (3).

  $\begin{array}{c}{c}^2=a^2-b^2\\ =16-4\\ =12\end{array}$

And,

  $c=\pm 2\sqrt{3}$ .

Thus the foci of the ellipse is,

  $\begin{array}{l}{F}_1=\left(2\sqrt{3,0}\right)\\ F_2=\left(-2\sqrt{3,0}\right)\end{array}$

The graph of ellipse is shown as,

  

Therefore, the solution is shown above.