Find the center, vertices, and foci of the ellipse with equation 2x2 + 7y2 = 14.

Answer 1

The standard form of  horizontal ellipse is [(x - h)²/a²] + [(y - k)²/b²)] = 1, when a² > b².

The ellipse equation 2x² + 7y² = 14.

The standard form for an ellipse is in a form = 1, So divide both sides of the equation by 14 to set it equal to 1.

2x²/14 + 7y²/14 = 14/14

(x²/7) + (y²/2) = 1

[(x - 0)²/7] + [(y - 0)²/2)] = 1.

Compare it to standard form of  horizontal ellipse is [(x - h)²/a²] + [(y - k)²/b²)] = 1, when a² > b².

Center (h, k) = (0, 0).

a = √7, b = √2, and c = √(a² - b²) = √(7 - 2) = √5.

Vertices : (h + a, k), (h - a, k) = (0 + √7, 0) and (0 - √7, 0) = (√7, 0) and (- √7, 0).

Foci : (h + c, k) and (h - c, k) = (0 + √5, 0) and (0 - √5, 0) = (√5, 0) and (- √5, 0).