x^2+3y^2−6x+6y+6=0. Find center,vertices,and the foci of the ellipse

Answer 1

The standard form for an ellipse is x²/a² + y²/b² = 1 (If the center is not at (0,0), then x and/or y get replaced by the unknown + or - a constant)

a and b are the "semi-major" and "semi-minor" axes---assuming that the longer axis of the ellipse is horizontal

The first thing to do is get your equation into this form....

x² + 3y² − 6x + 6y + 6 = 0

group the x and y terms, and factor out the constant for y:

x² − 6x ____ + 3(y² + 2y ____ ) + 6 = 0

Complete the square for each group:

x² − 6x + 9 + 3(y² + 2y + 1) + 6 - 9 - 3 = 0

Replace x and y polynomials by squares, and move all the constants to the right side:

(x - 3)² + 3(y + 1)² = 6

Divide both sides by 6:

(x - 3)² /6 + (y + 1)² /2 = 1

And put in standard form:

(x - 3)² / (√6)² + (y + 1)² / (√2)² = 1

So: a = 2.45 and b = 1.41

Note the offsets: These tell us that the center is at (3,-1)

The vertices are found by letting the y term go to zero---or by simply finding the x coordinates at 3 +/- a, a = 2.45, so x = 5.45 or 0.55

Finally, the distance from the center to the foci is commonly called "c", and is found from:

c² = a² - b²

c² = 6 + 2 = 8, so c = +/- 2.83

CHECK all this by using a graphing calculator or online tool to graph the original equation.