Question 826517
<pre>
y²-6y+48x²-3 = 0

You have to do algebraic operations on it to get it in the form:

{{{(x-h)^2/a^2}}}{{{""+""}}}{{{(y-k)^2/b^2}}}{{{""=""}}}{{{1}}}

in which case it will look like this {{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}},

or

{{{(x-h)^2/b^2}}}{{{""+""}}}{{{(y-k)^2/a^2}}}{{{""=""}}}{{{1}}}

in which case it will look like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}}

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 {{{1/2}}},
   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:

{{{48(x-0)^2/12}}}{{{""+""}}}{{{(y-3)^2/12}}}{{{""=""}}}{{{12/12}}}

{{{48(x-0)^2/12}}}{{{""+""}}}{{{(y-3)^2/12}}}{{{""=""}}}{{{1}}}

We must get the 48 off the top of the first fraction.
To do that we divide top and bottom by 48:

{{{expr(48/48)(x-0)^2/expr(12/48)}}}{{{""+""}}}{{{(y-3)^2/12}}}{{{""=""}}}{{{1}}} 

Simplify:

{{{1*(x-0)^2/expr(1/4)}}}{{{""+""}}}{{{(y-3)^2/12}}}{{{""=""}}}{{{1}}}

{{{(x-0)^2/expr(1/4)}}}{{{""+""}}}{{{(y-3)^2/12}}}{{{""=""}}}{{{1}}}

Now we can tell that it is an ellipse that looks like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}}

because the larger denominator is a² which equals 12, and the
smaller denominator is b² which equals {{{1/4}}}

So we compare it to:

{{{(x-h)^2/b^2}}}{{{""+""}}}{{{(y-k)^2/a^2}}}{{{""=""}}}{{{1}}}

(h,k) = (0,3) is the center, 

Since a² = 12 then a = &#8730;<span style="text-decoration: overline">12</span> = &#8730;<span style="text-decoration: overline">4*3</span> = 2&#8730;<span style="text-decoration: overline"></span> &#8776; 3.46

Since b² = {{{1/4}}} then b = {{{sqrt(1/4)}}} = {{{1/2}}}

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 &#8776; 3.46, units above and bellow the center, and the covertices
which are {{{1/2}}} of a unit right and left of the center:

{{{drawing(160,400,-2,2,-2,8,
circle(0,6.4641,0.12),circle(0,6.4641,0.11),circle(0,6.4641,0.09),circle(0,6.4641,0.07),circle(0,6.4641,0.05),circle(0,6.4641,0.03),circle(0,6.4641,0.01),
circle(0,-0.461,0.12),circle(0,-0.461,0.11),circle(0,-0.461,0.09),circle(0,-0.461,0.07),circle(0,-0.461,0.05),circle(0,-0.461,0.03),circle(0,-0.461,0.01),
circle(0.5,3,0.12),circle(0.5,3,0.11),circle(0.5,3,0.09),circle(0.5,3,0.07),circle(0.5,3,0.05),circle(0.5,3,0.03),circle(0.5,3,0.01),
circle(-0.5,3,0.12),circle(-0.5,3,0.11),circle(-0.5,3,0.09),circle(-0.5,3,0.07),circle(-0.5,3,0.05),circle(-0.5,3,0.03),circle(-0.5,3,0.01),

circle(0,3,0.09),circle(0,3,0.11),circle(0,3,0.12),

circle(0,3,0.07),circle(0,3,0.05),circle(0,3,0.03),circle(0,3,0.01),graph(160,400,-2,2,-2,8) )}}}

Then sketch in the ellipse (it's a tall skinny one!)

{{{drawing(160,400,-2,2,-2,8, arc(0,3,1,2sqrt(12)),graph(160,400,-2,2,-2,8),


circle(0,6.4641,0.12),circle(0,6.4641,0.11),circle(0,6.4641,0.09),circle(0,6.4641,0.07),circle(0,6.4641,0.05),circle(0,6.4641,0.03),circle(0,6.4641,0.01),
circle(0,-0.461,0.12),circle(0,-0.461,0.11),circle(0,-0.461,0.09),circle(0,-0.461,0.07),circle(0,-0.461,0.05),circle(0,-0.461,0.03),circle(0,-0.461,0.01),
circle(0.5,3,0.12),circle(0.5,3,0.11),circle(0.5,3,0.09),circle(0.5,3,0.07),circle(0.5,3,0.05),circle(0.5,3,0.03),circle(0.5,3,0.01),
circle(-0.5,3,0.12),circle(-0.5,3,0.11),circle(-0.5,3,0.09),circle(-0.5,3,0.07),circle(-0.5,3,0.05),circle(-0.5,3,0.03),circle(-0.5,3,0.01),

circle(0,3,0.09),circle(0,3,0.11),circle(0,3,0.12),

circle(0,3,0.07),circle(0,3,0.05),circle(0,3,0.03),circle(0,3,0.01) 


)}}} 

Edwin</pre>