The parts of the ellipse are:
In this case, we have:
• Vertices,: (3, 2) and (3, -6) → the vertices lie over the vertical line x = 3 → , the ellipse has the form of the ellipse at the ,right, in the graph,,
,
• Endpoints of minor axis,: (0, -2) and (6, -2).
In this case, the general equation of the ellipse is:
[tex]\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1.[/tex]
Where:
• (h, k) are the coordinates of the center of the ellipse,
,
• a and b are the legs.
From the graph at the right, we see that the equations for:
• the vertex are:
[tex]\begin{cases}\mleft(h,k+a\mright)=\mleft(3,2\mright), \\ (h,k-a)=(3,-6)\text{.}\end{cases}[/tex]
We see that h = 3. Summing the equations and solving for k and a, we get:
[tex]\begin{gathered} k+a=2,k-a=-6, \\ 2k=2-6=-4\Rightarrow k=-2, \\ -2+a=2\Rightarrow a=4. \end{gathered}[/tex]
So we have a = 4 and k = -2.
• the endpoints of the minor axis are:
[tex]\begin{gathered} (h+b,k)=(0,-2), \\ (h-b,k)=(-6,-2)\text{.} \end{gathered}[/tex]
Solving for b and replacing the value h = -3, we get:
[tex]h+b=0\Rightarrow b=-h\Rightarrow b=-(-3)=3.[/tex]
So we have b = 3.
The four parameters of the ellipse are:
• (h, k) = (3, -2),
,
• a = 4
,
• b = 3.
Replacing these values in the general equation of the ellipse, we get:
[tex]\begin{gathered} \frac{(x-3)^2}{3^2}+\frac{(y+2)^2}{4^2}=1, \\ \frac{(x-3)^2}{9^{}}+\frac{(y+2)^2}{16^{}}=1. \end{gathered}[/tex]
Answer
[tex]\frac{(x-3)^2}{9^{}}+\frac{(y+2)^2}{16^{}}=1[/tex]
