Question

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

Answer 1

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)

Answer 2

`{2}x^(2) + 8 {y}^(2) = 16 \\ \\ 1. \: {2x}^(2) = - 8 {y}^(2) + 16 \\ 2. \: {x}^(2) = - 4y^(2) + 8 \\ 3. \: x = \sqrt{ - 4y^(2) + 8} \\ x = - \sqrt{ - 4y^(2) + 8 } \\ \\ answer \\ x = \sqrt{ - 4y^(2) + 8 } \\ x = - \sqrt{ - 4y^(2) + 8}`