Question 1151192
<pre>{{{matrix(1,3,16x^2+y^2+64x-2y+67,""="",0)}}}

Swap the 2nd and 3rd terms on the left to get the terms together
for each letter:

{{{matrix(1,3,16x^2+64x+y^2-2y+67,""="",0)}}} 

Subtract the constant term from both sides:

{{{matrix(1,3,16x^2+64x+y^2-2y,""="",-67)}}}

{{{matrix(1,3,16x^2+64x+y^2-2y,""="",-67)}}}

Factor out 16 out of the first two terms, and 1 out of
the last two terms:

{{{matrix(1,3,16(x^2+4x)+1(y^2-2y),""="",-67)}}}

We complete the square inside the first parentheses:
1. Multiply the coefficient of x, which is 4, by one-half, getting 2.
2. Square 2, getting 4.
3. Add +4 at the end of the first parentheses.
4. Since the number in front of the parentheses is 16, adding 4 inside
   the parentheses amounts to adding 16∙4 or 64 to the left side, so
   we add 64 to the right side:

{{{matrix(1,3,16(x^2+4x+4)+1(y^2-2y),""="",-67+64)}}}

We complete the square inside the second parentheses:
1. Multiply the coefficient of y, which is -2, by one-half, getting -1.
2. Square -1, getting +1.
3. Add +1 at the end of the first parentheses.
4. Since the number in front of the parentheses is 1, adding 4 inside
   the parentheses amounts to adding 1∙1 or 1 to the left side, so
   we add 1 to the right side:

{{{matrix(1,3,16(x^2+4x+4)+1(y^2-2y+1),""="",-67+64+1)}}}

Next we factor the quadratics in the parentheses:

{{{matrix(1,3,16(x+2)(x+2)+1(y-1)(y-1),""="",-2)}}}

{{{matrix(1,3,16(x+2)^2+1(y-1)^2,""="",-2)}}}

This does not represent a conic because the left side is positive
and the right side is negative.  

We get 1 on the right side by dividing each term by -2

{{{matrix(1,3,16(x+2)^2/(-2)+1(y-1)^2/(-2),""="",(-2)/(-2))}}}

{{{matrix(1,3,16(x+2)^2/(-2)+1(y-1)^2/(-2),""="",1)}}}

We divide top and bottom of the first fraction by 16

{{{matrix(1,3,(x+2)^2/(-2/16)+1(y-1)^2/(-2),""="",1)}}}

{{{matrix(1,3,(x+2)^2/(-1/8)+1(y-1)^2/(-2),""="",1)}}}

{{{matrix(1,3,(x+2)^2/(-1/8)+(y-1)^2/(-2),""="",1)}}}

If this were a real conic, there would not be a negative number
on the bottom of either term, but only positive numbers.

Did you copy the problem wrong?

Edwin</pre>