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


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
[tex]\begin{gathered} \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1 \\ a>b \end{gathered}[/tex]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
[tex]vertices:(0,-16+20)=(0,4)\text{ and }(0,-16-20)=(0,-36)[/tex]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
[tex]\begin{gathered} \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1 \\ center:(h,k),a>b \\ Vertices:(h,k\pm a) \end{gathered}[/tex]Thus, in our case,
[tex]\begin{gathered} \frac{(y+13)^2}{12^2}-\frac{x^2}{5^2}=1 \\ \Rightarrow vertices:(0,-13+12)=(0,-1),(0,-13-12)=(0,-25) \\ \end{gathered}[/tex]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).
C)
In general, the foci of an ellipse/hyperbola are given by the formulas below
[tex]\begin{gathered} Hyperbola \\ focal\text{ distance: }c^2=a^2+b^2 \\ Ellipse \\ c^2=a^2-b^2 \end{gathered}[/tex]Then,
[tex]\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}[/tex][tex]\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}[/tex]