Find the center, vertices, foci, and eccentricity of the ellipse. 6x^2 + 2y^2 + 30x − 18y + 54 = 0

Answer 1

First rearrange the equation so x terms are next to each other and y terms are next to each other

6x² + 30x + 2y² - 18y + 54 = 0

Subtract the 54 over

6x² + 30x + 2y² - 18y = -54

Next do complete the square with the x terms.

Rewrite 6x² + 30x as 6(x² + 5x)

To complete the square, we add our second coefficient (5) divided by two and squared.

So we add (5/2)² or 25/4

Now we have 6(x² + 5x + 25/4). Distribute the six back in.

6x² + 30x + 75/2. So this is our square polynomial that's equal to 6(x + 5/2)²

From the original equation add 75/2 to both sides.

6x² + 30x + 75/2 + 2y² - 18y = -54 + 75/2

6(x + 5/2)² + 2y² - 18y = -54 + 75/2

Repeat by doing complete the square with the y terms. You should get:

6(x + 5/2)² + 2(y - 9/2)²= -54 + 75/2 + 81/2

6(x + 5/2)² + 2(y - 9/2)² = 24

(x + 5/2)² /4 + (y - 9/2)² /12 = 1

a = 2√3, b = 2

c² = a² - b² = 8, c = 2√2

The center is at (-5/2, 9/2)

Since the greater denominator is under the y term, the foci and vertices lie vertically.

Foci: (-5/2, 9/2 ± c)

Vertices: (-5/2, 9/2 ±a)

Eccentricity = c/a