Question

Find the center, vertices, and foci of the ellipse with the given equation. (x+4)2/400 + (y+3)2/144 = 1

Answer 1

`\textit{ellipse, horizontal major axis} \\\\ \cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad √( a ^2- b ^2) \end{cases} \\\\[-0.35em] ~\dotfill`

`\cfrac{(x+4)^2}{400}~~ + ~~\cfrac{(y+3)^2}{144}~~ = ~~1\implies \cfrac{[x-(-4)]^2}{20^2}~~ + ~~\cfrac{[y-(-3)]^2}{12^2}~~ = ~~1 \\\\\\ \begin{cases} h=-4\\ k=-3\\ a=20\\ b=12 \end{cases}\qquad \qquad c=√(20^2-12^2)\implies c=√(256)\implies \boxed{c=16} \\\\[-0.35em] ~\dotfill`

`\stackrel{center}{(-4,-3)}\qquad vertices \begin{cases} (\stackrel{-4-20}{-24}~~,~~-3)\\\\ (\stackrel{-4+20}{16}~~,~~-3) \end{cases}\qquad foci \begin{cases} (\stackrel{-4-16}{-20}~~,~~-3)\\\\ (\stackrel{-4+16}{12}~~,~~-3) \end{cases}`