Question 1204202
<pre>

{{{36x^2+25y^2-288x+200y+76=0}}}

Get the x terms together and the y-terms together and
the constant term on the right

{{{36x^2-288x+25y^2+200y=-76}}}

Factor the coefficient of the squared terms out of the first two terms
and the last two terms on the left:

{{{36(x^2-8x)+25(y^2+8y)=-76}}}

We want to add a positive number inside each parentheses to make
it factorable as a perfect square.  To find that number

In the first parentheses,
Half the coefficient of x, which is -8, getting -4, then square
-4 getting +16 and add that inside the first parentheses on the right. Since 
there is a coefficient of 36, adding +16 inside the first parentheses is
the equivalent of adding 36x16 or 576 to the left side. So we must add
576 to the right side as well:

{{{36(x^2-8x+16)+25(y^2+8y)=-76+576}}}

In the second parentheses,
Half the coefficient of y, which is +8, getting +4, then square
+4 getting +16 and add that inside the second parentheses on the right. Since 
there is a coefficient of 25, adding +16 inside the second parentheses is
the equivalent of adding 25x16 or 400 to the left side. So we must add
400 to the right side as well:

{{{36(x^2-8x+16)+25(y^2+8y+16)=-76+576+400}}}

Now we factor each parentheses and combine the numbers on the right side.

{{{36(x-4)(x-4)+25(y+4)(y+4)=900}}}

We notice that they factored into perfect squares, so perhaps you can skip
the preceding step, and have gone straight to

{{{36(x-4)^2+25(y+4)^2=900}}}
 
Now we must get 1 on the right side, so we divide through by 900

{{{36(x-4)^2/900^""+25(y+4)^2/900^""=900^""/900^""}}}

The coefficients of the squares of binomials must be 1, so divide
the top and bottom of the first term by 36 and the second by 25.

{{{(x-4)^2/25^""+(y+4)^2/36^""=1}}}

{{{(x-4)^2/5^2+(y+4)^2/6^2=1}}} compare to {{{(x-h)^2/b^2+(y-k)^2/a^2=1}}}

That's a vertical ellipse (looks like the number "0") because the larger
denominator a<sup>2</sup> is under the term in y.  

It has center (h,k)=(4,-4). 

It has semi-major axis a=6, semi-minor axis b=5.

Vertices (4,-10) and (4,2) and co-vertices (-1,-4) and (9,-4).

{{{drawing(400,3200/7,-3,11,-12,4, arc(4,-4,10,-12),circle(4,-4,.1),

grid(1))}}}

Edwin</pre>