Answer:
[tex]\dfrac{x^2}{144}+\dfrac{(y-3)^2}{225}=1[/tex]
Step-by-step explanation:
Since the x-coordinates of the foci are the same, the ellipse is vertical.
The general equation of a vertical ellipse is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a vertical ellipse}}\\\\\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\\\\\textsf{where:}\\\phantom{ww}\bullet \textsf{$2b$ is the major axis.}\\\phantom{ww}\bullet \textsf{$2a$ is the minor axis.}\\\phantom{ww}\bullet \textsf{$(h,k)$ is the center.}\\ \phantom{ww}\bullet \textsf{$(h,k\pm b)$ are the vertices.}\\ \phantom{ww}\bullet \textsf{$(h,k\pm c)$ are the foci where $c^2=b^2-a^2$}\end{array}}[/tex]
The center (h, k) of the ellipse is the midpoint of the foci. Therefore:
[tex]h = 0[/tex]
[tex]k = \dfrac{12+(-6)}{2}=\dfrac{6}{2}=3[/tex]
The major axis of an ellipse is the longest diameter that passes through the center of the ellipse and intersects both foci.
Given that the major axis (2b) is 30, then:
[tex]2b=30\\\\\\b=\dfrac{30}{2}\\\\\\b=15[/tex]
Given that the foci are (0, 12) and (0, -6), and h = 0 and k = 3, we can determine the value of c using the foci formula:
[tex](h,k\pm c)=(0, 3\pm c)\\\\3\pm c=12,-6\\\\\pm c=(12-3),(-6-3)\\\\\pm c=9,-9\\\\\pm c=\pm9\\\\c=9[/tex]
Now that we have the values of b and c, we can find the value of a² by using the formula c² = b² - a²:
[tex]9^2=15^2-a^2\\\\a^2=15^2-9^2\\\\a^2=225-81\\\\a^2=144[/tex]
Substitute h = 0, k = 3, a² = 144 and b = 15 into the general equation:
[tex]\dfrac{(x-0)^2}{144}+\dfrac{(y-3)^2}{15^2}=1\\\\\\\dfrac{x^2}{144}+\dfrac{(y-3)^2}{225}=1[/tex]
Therefore, the equation of the ellipse with foci (0 ,12) and (0, -6), and a major axis with a length of 30 is:
[tex]\Large\boxed{\boxed{\dfrac{x^2}{144}+\dfrac{(y-3)^2}{225}=1}}[/tex]