9x2+36x+4y2-24y+3-6=0 can be written so as to complete two squares:
9(x2+4x+4-4)+4(y2-6y+9-9)+36=0,
9((x+2)2-4)+4((y-3)2-9)+36=0,
9(x+2)2-36+4(y-3)2-36+36=0,
9(x+2)2+4(y-3)2-36=0,
by dividing through by 36: (x+2)2/22+(y-3)2/32=1 is the standard equation of an ellipse with centre (-2,3). These coordinates are taken from the squared terms x+2 and y-3. To make this the origin or centre (0,0), x+2=0, so x=-2 and y-3=0 so y=3. Now we look at the square roots of the denominators. These tell us the lengths of the semi-axes. For the x-axis we have 2 and for the y-axis we have 3. So 3 is the semi-major axis, being longer than 2, the semi-minor axis. Therefore, the ellipse is taller than it is wide and the lengths of the two axes are 6 and 4.
But there's something else about this ellipse. The length of the semi-minor axis is 2 and the centre has an x coordinate of -2, which means the ellipse touches the y-axis (tangent to it). And the same coincidence applies to the y coordinate, and the ellipse touches the x-axis.
