find the center and vertices of the ellipce given by the equation?

Question

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

Answer 1

4x^2-24x+36+y^2-4y+4-36-4+36=0
(2x-6)^2+(y-2)^2=2^2
this is the equation of an ellipse with centre(3,2)
vertices (3,0,(3,4),(2,2),(4,2)

Answer 2

4x^2+y^2-24x-4y+36=0
4x^-24x+36 +y^2 -4y =0
4(x^2-6x+9) + y^2-4y +4 =4
4(x-3)^2 + (y-2)^2= 4
(x-3)^2/1 + (y-2)^2)/4 = 1
This is the equation of an ellipse with center at (3,2)
Since a =1 (a is the sqrt of the denominator of (X-3)^2, the vertices are at (-a,2) and (a,2) = (-1.2) and (1,2) [The 2 comes from the y-value of the center of the ellipse]