Question 599384
(x-2)^2 over 4 +(y-3)^2 over 16=1
(x-2)^2/4+(y-3)^2/16=1
This is an equation of an ellipse with vertical major axis.
Its standard form: (x-h)^2/b^2+(y-k)^2/a^2=1, a>b, (h,k)=(x,y) coordinates of center
..
For given equation:
Center: (2,3)
a^2=16
a=√16=4
length of vertical major axis=2a=8
b^2=4
b=2
length of minor axis=2b=4
y=±(16-4(x-2)^2)^.5+3

see graph below
{{{ graph( 300, 300, -10, 10, -10, 10, (16-4(x-2)^2)^.5+3,-(16-4(x-2)^2)^.5+3) }}}