Question

In a solar system… Question and rubric linked below. Thanks for your help!

Answer 1

A) Notice that the trajectory of comet E corresponds to an ellipse whose major axis is parallel to the y-axis.

In general, the equation of an ellipse centered at (h,k) is

`\begin{gathered} ((x-h)^2)/(b^2)+((y-k)^2)/(a^2)=1 \\ a>b \end{gathered}`

Then, in our case, the ellipse is centered at (0,-16), its semimajor axis length (a) is sqrt(400)=20, and its semiminor axis length is sqrt(144)=12.

Therefore, the vertices of the ellipse are

`vertices:(0,-16+20)=(0,4)\text{ and }(0,-16-20)=(0,-36)`

The answers to part A) are (0,-36) and (0,4).

B)

On the other hand, the equation of Comet H corresponds to a hyperbola whose transverse axis is on the y-axis. Therefore, its minimum distance to (0,0) is given by one of its vertices.

Calculate the vertices of the trajectory of comet H as shown below

`\begin{gathered} ((y-k)^2)/(a^2)-((x-h)^2)/(b^2)=1 \\ center:(h,k),a>b \\ Vertices:(h,k\pm a) \end{gathered}`

Thus, in our case,

`\begin{gathered} ((y+13)^2)/(12^2)-(x^2)/(5^2)=1 \\ \Rightarrow vertices:(0,-13+12)=(0,-1),(0,-13-12)=(0,-25) \\ \end{gathered}`

Comet H passes through (0,-1) which is at 1 unit from the origin.

On the other hand, the closest position of comet E from the sun is at its vertex (0,4).

The answer to part B is that comet H is closer to the sun at its maximum approximation point.

C)

In general, the foci of an ellipse/hyperbola are given by the formulas below

`\begin{gathered} Hyperbola \\ focal\text{ distance: }c^2=a^2+b^2 \\ Ellipse \\ c^2=a^2-b^2 \end{gathered}`

Then,

`\begin{gathered} Comet\text{ E} \\ c^2=400-144=256 \\ \Rightarrow c=16 \\ \Rightarrow foci:(0,-16+16),(0,-16-16) \\ \Rightarrow foci:(0,0),(0,-32) \end{gathered}`
`\begin{gathered} Comet\text{ H} \\ c^2=144+25=169 \\ \Rightarrow c=13 \\ \Rightarrow foci:(0,-13+13),(0,-13-13) \\ \Rightarrow foci:(0,0),(0,-26) \end{gathered}`

Thus, (0,0) (sun's position) is a focus of both comets.