[tex] {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} [/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)
