Find center, foci, and vertices.

Question

3x^2+4y^2-36x+24y+132=0

Answer 1

Looking at the equation we can tell it will be an ellipse because of the coefficients of x² and y² are not the same but they are the same sign (so not a hyperbola).  It's easiest to find the critical points like the center foci and vertices if it's written in the form: (x-h)²/(a²)+(y-k)²/(b²) = 1

 

First we reorganize a little:

3x²-36x+4y²+24y = -132

 

Then we try to complete the square for x and y.  We'll start with y:

 

3x²-36x + 4(y²+6x +9) = -132+36 (since we factored by 4)

3x²-36x + (y+3)² = -96

 

now for x:

 

3(x²-12x+36)+4(y+3)²=-96+108 (since I factored out a 3, we added 108 not 36)

3(x-6)²+4(y+3)²=12

 

Finally, we divide by 12 to get the right hand side equal to 1:

(x-6)²/4 + (y+3)²/3 = 1

 

Now we can read off the points you need:  

 

The center is shifted from (0,0) to (6,-3) (this is from the h, k).  

 

The major axis is√4 = 2 left and right from the center (since the larger number is under the x term we can tell it will be longer in the horizontal direction).  So the major vertices are: (4,-3)) and (8,-3).  The minor vertices are: (6,-3-√3) (6,3+√3)).  The foci distance "c" is found by the ellipse relation: c²=a²-b².  4-3 = 1. so c = √1 = 1.  Finally the foci points are a distance c from the center in the major axis direction:  F1 = (5,-3) and F2 = (7,-3)

Answer 2

Put the ellipse into standard form. An ellipse with center (h,k), semi-major axis a, and semi-minor axis b has an equation of the form:

(x - h)²/a² + (y - k)²/b² = 1
If the major axis runs horizontally, and

(x - h)²/b² + (y - k)²/a² = 1
If the major axis runs vertically.

Now to the equation at hand.

3x² - 36x + 4y² + 24y + 132 = 0

First complete the squares.

3(x² - 12x) + 4(y² + 6y) + 132 = 0
3(x² - 12x+(-12/2)² – (-12/2)²) + 4(y² + 6y+(6/2)² – (6/2)²) + 132 = 0
3(x²-12x+36) – 3(36) + 4(y²+6y+9) – 4(9) + 132 = 0
3(x-6)² + 4(y+3)² = 3(36) + 4(9) -132
3(x-6)² + 4(y+3)² = 12

Now set equal to one by dividing by 12.
(x-6)²/4 + (y+3)²/3 = 1

a² = 4 and b² = 3
a = 2 and b = √3

The center of the ellipse is (h,k) = (6,-3).
The major vertices are (h-a,k) and (h+a,k) or
(6-2,-3) and (6+2,-3) which is (4,-3) and (8,-3).

c = √(a² - b²) = √(4 - 3) = √1 = 1

The foci are (h-c,k) and (h+c,k) or
(6-1,-3) and (6+1,-3) which is (5,-3) and (7,-3).