Question 544552
{{{(x+2)^2/9 + (y+2)^2/4 = 1 }}}
Think of it this way:
{{{x^2+y^2=1}}} represents a circle centered at (0,0) with radius 1.
(You can see that the formula says that the distance to the center is 1).
{{{(x+2)^2 + (y+2)^2 = 1 }}} would represent a circle centered at (-2,-2) with radius 1.
Stretching the drawing in the x and y directions to the same extent you would get a bigger circle with the same center:
{{{(x+2)^2/9 + (y+2)^2/9 = 1 }}} <--> {{{(x+2)^2 + (y+2)^2 = 9 }}} <--> {{{(x+2)^2 + (y+2)^2 = 3^2 }}} would represent a circle centered at (-2,-2) with radius 3. 
{{{(x+2)^2/4 + (y+2)^2/4 = 1 }}} <--> {{{(x+2)^2 + (y+2)^2 = 4 }}} <--> {{{(x+2)^2 + (y+2)^2 = 2^2 }}} would represent a circle centered at (-2,-2) with radius 2.
Your ellipse is a stretched circle, symmetrical about axes {{{x=-2}}} and {{{y=-2}}}. It's only that the guy stretching in the x direction was pulling harder.
{{{x=-2}}} (substitute and see) at {{{y=2}}} and {{{y=-2}}}.
It crosses axis {{{y=-2}}} at {{{x= 3}}} and {{{x= - 3}}}.
{{{ graph( 300, 200, -5.3, 1.3, -4.2, 0.2, 2sqrt(1-(x+2)^2/9)-2, -2sqrt(1-(x+2)^2/9)-2) }}}