Explanation:
The equation of comet E is given below as
[tex]\frac{(y+16)^2}{400}+\frac{x^2}{144}=1[/tex]The equation of comet H is modelled below as
[tex]\frac{(y+13)^2}{144}-\frac{x^2}{25}=1[/tex]Given, that the Sun is located at the origin.
This is an equation of an ellipse. So, the path traveled by Comet e is an elliptic path.
Comparing with the Standard equation of ellipse: below,we will have
[tex]\frac{(y-k)^2}{b^2}+\frac{(x-h)^2}{a^2}[/tex]Whise center is
[tex]\begin{gathered} center=(h,k) \\ vertices=(h\pm a,0) \end{gathered}[/tex]By comparing coefficient, we will have
[tex]\begin{gathered} a^2=144 \\ a=12 \\ b^2=400 \\ b=20 \\ k=-16 \\ h=0 \\ (h,k)=(0,-16) \end{gathered}[/tex]Hence,
The vertex of comet E will be
[tex]\begin{gathered} (h\pm a,0) \\ (0+12,0),(0-12,0) \\ (12,0),(-12,0) \end{gathered}[/tex]The vertex of comet E is
[tex](12,0),(-12,0)[/tex]Part C:
It is the foci forboth comets
[tex]\begin{gathered} cometE: \\ c=\sqrt{b^2-a^2} \\ c=\sqrt{400-144} \\ c=\sqrt{256} \\ c=16 \\ \\ For\text{ comet F:} \\ c=\sqrt{a^2-b^2} \\ c=\sqrt{144+25} \\ c=\sqrt{169} \\ c=13 \end{gathered}[/tex]Hence,
The foci will be
[tex]\begin{gathered} (16\pm16,0) \\ cometE \\ (0,0),(32,0) \\ cometF: \\ (13\pm13,0) \\ (0,0),(26,0) \end{gathered}[/tex]