Please use proper English. I cannot understand what you are asking if you don't use a proper sentence.
I will complete the square.
12(x^2 - 24x +144) + 4(y^2 - y + 1/4) = -1+12(144)+4(1/4)
12(x-12)^2 + 4(y - 1/2)^2 = 1728
divide by 1728
(x-12)^2/144 + (y-1/2)^2/432 = 1
from there we know a = 12 , b = 12√3
find c, and take it from there
finding the standard form the center eccentricity vertices foci and minor axis endpoints of the equation 12x^2+4y^2-24x-4y+1=0
2 answers
I factored incorrectly, let me try this again
12(x^2 - 2x + 1) + 4(y^2 - y + 1/4) = -1 + 12 - 1
12(x-1)^2 + 4(Y-1/2)^2 = 12
(x-1)^2 /1 + (y-1/2)^2 /3 = 1
etc
12(x^2 - 2x + 1) + 4(y^2 - y + 1/4) = -1 + 12 - 1
12(x-1)^2 + 4(Y-1/2)^2 = 12
(x-1)^2 /1 + (y-1/2)^2 /3 = 1
etc