Question 688516
For the equation of the ellipse, find the coordinates of the CENTER, FOCI, and VERTICES. THEN GRAPH. 
4x^2+9y^2-8x+36y+4=0
complete the square
4x^2-8x+9y^2+36y+4=0
4(x^2-2x+1)+9(y^2+4y+4)=-4+4+36
4(x-1)^2+9(y+2)^2=36
(x-1)^2/9+(y+2)^2/4=1
This is an equation of an ellipse with horizontal major axis.
Its standard form: {{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}, a>b, (h,k)=(x,y) coordinates of center
For given ellipse:
center:(1,-2)
a^2=9
a=√9=3
vertices: (1±a,-2)=(1±3,-2)=(4,-2) and (-2,-2)
b^2=4
b=√4=2
c^2=a^2-b^2=9-4=5
c=√5≈2.24
Foci:(1±c,-2)=(1±2.24,-2)=(3.24,-2) and (-1.24,-2)
see graph below:
y=±(4-(4(x-1)^2/9))^.5-2
{{{ graph( 300, 300, -10, 10, -10, 10, (4-(4(x-1)^2/9))^.5-2,-(4-(4(x-1)^2/9))^.5-2) }}}