y²-6y+48x²-3 = 0
You have to do algebraic operations on it to get it in the form:
in which case it will look like this 
,
or
in which case it will look like this 
We won't know which it is until later, when we can see which
denominator is larger, for the larger denominator will be "a²" and
the smaller denominator will be "b²".
y²-6y+48x²-3 = 0
Put the x term first, then the y terms second on the left
of the equation and isolate the number on the right:
48x²+y²-6y = 3
Since there is no x-term, all we have to do is write the
term 48x² as 48(x-0)².
48(x-0)²+y²-6y = 3
Since there is a y-term we must complete the square on
y²-6y:
To the side, we
1. Multiply the coefficient of y, which is -6, by 
,
getting -3.
2. Square the result of step 1 (-3)² = 9
3. Add +9 to both sides of the equation
48(x-0)²+y²-6y+9 = 3+9
Then we factor y²-6y+9 as (y-3)(y-3) and then as (y-3)²
and replace y²-6y+9 by (y-3)². Combine the numbers on
the right side:
48(x-0)²+(y-3)² = 12
Then we divide through by 12 to get 1 on the right:
We must get the 48 off the top of the first fraction.
To do that we divide top and bottom by 48:
Simplify:
Now we can tell that it is an ellipse that looks like this
because the larger denominator is a² which equals 12, and the
smaller denominator is b² which equals 
So we compare it to:
(h,k) = (0,3) is the center,
Since a² = 12 then a = √12 = √4*3 = 2√ ≈ 3.46
Since b² = then b = 
=
a = the distance from the center (0,3) to the vertices, and b = the distance
from the center to the covertices. So we plot the center, the vertices,
which are a ≈ 3.46, units above and bellow the center, and the covertices
which are of a unit right and left of the center:
Then sketch in the ellipse (it's a tall skinny one!)
Edwin
