How do you put #12x^2+3y^2-30y+39=9# in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

1 Answer
Dec 30, 2017

Answer:

#(4x^2)/15 + (y-5)^2/15 = 1; " center": (0, 5)#
vertices: #(0, 5+sqrt(15)), (0, 5- sqrt(15))#
endpoints (co-vertices): #(-sqrt(15)/2, 5), (sqrt(15)/2, 5)#
foci: #(0, 5+(3sqrt(5))/2), (0, 5-(3sqrt(5))/2)#; #e = sqrt(3)/2#

Explanation:

Given: #12x^2 + 3y^2-30y+39 = 9#

To Find the standard form of the equation , use the completing of the square technique :

Group both the #x#-terms and the #y#-terms together on the left-side of the equation and all the other constants on the right side:
#12x^2 + (3y^2-30y) = -30#

Factor if needed: #" "12x^2 + 3(y^2-10y) = -30#

We only need to use completing of the square with the #y#-terms. Half the #-10y " term "-> -5 " and add the " 3(-5)^2# to the right side that was added to the left side when we completed the square:

#" "12x^2 + 3(y-5)^2 = -30 + 3(5)^2#

#" "12x^2 + 3(y-5)^2 = 45#

Divide everything by #45# to make the equation #= 1#:

Standard form of the equation:

#" "(4x^2)/15 + ((y-5)^2)/15 = 1 " or " x^2/(sqrt(15)/2)^2 + (y-5)^2/(sqrt(15)^2) = 1#

Because we have a "sum" in the equation, we have an ellipse . In an ellipse #a# is always the major axis (longest). This means we have a Vertical Major Axis ellipse: #(x-h)^2/b^2 + (y-k)^2/a^2 = 1#

Where #" **center** "= (h, k) = (0, 5)#

#a = sqrt(15), " "b = sqrt(15)/2#
#c = sqrt(a^2 - b^2) = sqrt(15-15/4) = sqrt(45/4) = (3sqrt(5))/2#

Vertices: #(h, k+a), (h, k-a): (0, 5 +sqrt(15)), (0, 5-sqrt(15))#

Endpoints (co-vertices): #(h-b, k), (h+b, k): (-sqrt(15)/2, 5), (sqrt(15)/2, 5)#

Foci: #(h, k+c), (h, k-c): (0, 5+(3sqrt(5))/2), (0, 5-(3sqrt(5))/2)#

Eccentricity:
#varepsilon = c/a = ((3sqrt(5))/2)/sqrt(15)=(3sqrt(5))/2 * 1/sqrt(15) *sqrt(15)/sqrt(15) = (3sqrt(75))/30 = sqrt(75)/10#

#varepsilon =sqrt(3)/2#