Respuesta :
[tex]\bf \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 \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf 2x^2+8y^2=16\implies \cfrac{2x^2+8y^2}{16}=1\implies \cfrac{2x^2}{16}+\cfrac{8y^2}{16}=1\implies \cfrac{x^2}{8}+\cfrac{y^2}{2}=1 \\\\\\ \cfrac{(x-0)^2}{(\sqrt{8})^2}+\cfrac{(y-0)^2}{(\sqrt{2})^2}=1\qquad \begin{cases} a=\sqrt{8}\\ b=\sqrt{2}\\ c=\sqrt{a^2-b^2}\\ \qquad \sqrt{6} \end{cases}~\hfill \begin{array}{llll} \stackrel{\textit{vertices}}{(\pm \sqrt{8},0)}\qquad \stackrel{\textit{foci}}{(\pm \sqrt{6},0)}\\\\ ~\hfill \stackrel{\textit{center}}{(0,0)}~\hfill \end{array}[/tex]
Answer:
Step-by-step explanation:
2x² + 8y² = 16
divide both sides of equation by the constant
2x²/16 + 8y²/16 = 16/16
x²/8 + y²/2 = 1
x² has a larger denominator than y², so the ellipse is horizontal.
General equation for a horizontal ellipse:
(x-h)²/a² + (y-k)²/b² = 1
with
a² ≥ b²
center (h,k)
vertices (h±a, k)
co-vertices (h, k±b)
foci (h±c, k), c² = a²-b²
Plug in your equation, x²/8 + y²/2 = 1.
(x-0)²/(2√2)² + (y-0)²/(√2)² = 1
h = k = 0
a = 2√2
b = √2
c² = a²-b² = 6
c = √6
center (0,0)
vertices (0±2√2,0) = (-2√2, 0) and (2√2, 0)
co-vertices (0, 0±√2) = (0, -√2) and (0, √2)
foci (0±√6, 0) = (-√6, 0) and (√6, 0)
