Question

finding the standard form the center eccentricity vertices foci and minor axis endpoints of the equation 12x^2+4y^2-24x-4y+1=0

Answer 1

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 <b>+144</b>) + 4(y^2 - y + <b>1/4</b>) = -1<b>+12(144)</b>+<b>4(1/4)</b>
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

Answer 2

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