Question

What is the standard form equation of the ellipse that has vertices (-2,-8) and (-2,2) and has foci (-2,-7) and (-2,1) ?​

Answer 1

if we plot those 4 points, the two vertices and the two foci, we'll notice that they're all colinear vertically, namely making a vertical line, which in short means is a vertical ellipse, which means the "a" component will go under the fraction with the "y" above.

the half-way between the vertices lands on (-2, -3), thus that is our center.

the distance from one vertex to the other is 10 units, meaning a = 5.

the distance from the center to either foci is 4 units, meaning c = 4.

what's "b" anyway?

$\bf \textit{ellipse, vertical major axis} \\\\ \cfrac{(x- h)^2}{ b^2}+\cfrac{(y- k)^2}{ a^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h = -2\\ k = -3\\ a = 5\\ c = 4 \end{cases}\implies \cfrac{[x-(-2)]^2}{b^2}+\cfrac{[y-(-3)]^2}{5^2}=1 \\\\[-0.35em] ~\dotfill$

$\bf c = \sqrt{a^2-b^2}\implies 4=\sqrt{5^2-b^2}\implies 4^2=5^2-b^2\implies \implies b^2+4^2=5^2 \\\\\\ b^2=5^2-4^2\implies b^2=25-16\implies b^2=9\implies b=\sqrt{9}\implies \boxed{b = 3} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[x-(-2)]^2}{3^2}+\cfrac{[y-(-3)]^2}{5^2}=1\implies \cfrac{(x+2)^2}{9}+\cfrac{(y+3)^2}{25}=1$

Check the picture below.