0

# Find center, foci, and vertices.

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

### 2 Answers by Expert Tutors

Peter Y. | Peter's Profile: Math TutoringPeter's Profile: Math Tutoring
4.9 4.9 (37 lesson ratings) (37)
0
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)
Xavier H. | Excellent Math and Science TutorExcellent Math and Science Tutor
5.0 5.0 (13 lesson ratings) (13)
0
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).