Answer:
[tex]\dfrac{(x+5)^2}{5^2}+\dfrac{(y-6)^2}{8^2}=1[/tex]
Step-by-step explanation:
The vertices of an ellipse are at (-5, -2) and (-5, 14). Given that their x-coordinates are the same, this means that the major axis is parallel with the y-axis, and so 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) is located at the midpoint of the vertices:
[tex](h,k)=\left(\dfrac{-5-5}{2},\dfrac{14-2}{2}\right)\\\\\\(h,k)=\left(\dfrac{-10}{2},\dfrac{12}{2}\right)\\\\\\(h,k)=\left(-5,6\right)\\\\\\[/tex]
To find the value of b (major radius), substitute h = -5, k = 6 and the coordinates of one of the vertices into the vertex formula and solve for b:
[tex](h,k+b)=(-5,14)\\\\(-5,6+b)=(-5,14)\\\\6+b=14\\\\b=8[/tex]
Therefore, the equation so far is:
[tex]\dfrac{(x-(-5))^2}{a^2}+\dfrac{(y-6)^2}{8^2}=1\\\\\\\dfrac{(x+5)^2}{a^2}+\dfrac{(y-6)^2}{8^2}=1[/tex]
To find the value of a (minor radius), substitute the coordinates of the point (0, 6) that lies on the ellipse into the equation and solve for a:
[tex]\dfrac{(0+5)^2}{a^2}+\dfrac{(6-6)^2}{8^2}=1\\\\\\\dfrac{(5)^2}{a^2}+\dfrac{(0)^2}{8^2}=1\\\\\\\dfrac{25}{a^2}=1\\\\\\a^2=25\\\\\\a=5[/tex]
Therefore, the equation of the ellipse with vertices at (-5, -2) and (-5, 14), and that passes through point (0, 6), is:
[tex]\Large\boxed{\boxed{\dfrac{(x+5)^2}{5^2}+\dfrac{(y-6)^2}{8^2}=1}}[/tex]
