Question

Find the center, vertices, and foci of the ellipse with equation 2x2 + 8y2 = 16.

Answer 1

`\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 √( a ^2- b ^2) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\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}{(√(8))^2}+\cfrac{(y-0)^2}{(√(2))^2}=1\qquad \begin{cases} a=√(8)\\ b=√(2)\\ c=√(a^2-b^2)\\ \qquad √(6) \end{cases}~\hfill \begin{array}{llll} \stackrel{\textit{vertices}}{(\pm √(8),0)}\qquad \stackrel{\textit{foci}}{(\pm √(6),0)}\\\\ ~\hfill \stackrel{\textit{center}}{(0,0)}~\hfill \end{array}`

Answer 2

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)