Question 612903
{{{ x^2-2x-4y^2-8y-7=0}}}
Because the squared terms have opposite signs, this equation is the equation of a hyperbola. The general standard form for a hyperbola is:
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}
The numerators are the squares of binomials and to get these we "complete the square" for both the x terms and the y terms. The procedure is:<ol><li>"Move" the constant term to one side of the equation (with all the variable terms on the other side).</li><li>Rearrange the order of the terms on the variable side so that the x terms are together and the y terms are together.</li><li>For each variable, factor out the coefficient of the squared term (if it is not a 1) from both terms. NOTE: Pay attention to "-" signs in front of the squared terms!</li><li>For each variable, find half of the first power term.</li><li>For each variable, find the square of the half you found in the previous step.</li><li>Add the squares found in the previous step to both sides of the equation. NOTE: This can be much trickier than it may sound (as we will find out shortly).</li><li>And last, rewrite as squares of binomials. The general form is:
"(x + the half from step 4) squared" and "(y + the half from step 4) squared"</li></ol>Let's see this in action:
1. Move the constant term.
Adding 7 to each side...
{{{ x^2-2x-4y^2-8y=7}}}
2. Rearrange the terms...
Your x terms and y terms are already together.
3. Factor out coefficients of the squared terms.
Your {{{x^2}}} has a coefficient of 1 so we can skip this. Your {{{y^2}}} term has a coefficient of -4:
{{{ x^2-2x-4(y^2+2y)=7}}}
Note how factoring out the -4 changes the sign of the +8y!
4. The 1st power term for x is -2x, The coefficient is -2 and half of this is -1. The 1st power term of  y is 2y. The coefficient is 2 and half of this is 1.
5. Find the squares of the halves.
Both -1 squared and 1 squared is 1.
6. Add the squares to each side.
This is the hardest part to get right. This is how the left side should look;
{{{ x^2-2x+ 1 -4(y^2+2y+ 1)}}}
Note where the squares went, especially the square for the y terms. It is inside the parentheses! The 1 we added to the x terms is not in parentheses and so it is just a 1. But the 1 for the y terms is inside the parentheses. Because it is inside the parentheses and because there is a -4 in front of the parentheses, this "1" is really a -4! So when we add the same thing to the right side that we added to the left side we need to add a 1 and a -4!
{{{ x^2-2x+ 1 -4(y^2+2y+ 1) = 7 + 1 + (-4)}}}
Simplifying the right side we get:
{{{ x^2-2x+ 1 -4(y^2+2y+ 1) = 4}}}
7. Rewrite as squares of binomials, using the "halves" from step 4:
{{{ (x+(-1))^2 -4(y+1)^2 = 4}}}<br>
After all that we have completed the squares. For the standard form we need:<ol><li>A 1 on the right side</li><li>No coefficients in front of the completed squares</li><li>Perfect square denominators under the completed squares.</li><li>Rewrite the completed squares as subtractions.</li></ol>
To get the 1 on the right, divide both sides by the 4 that is there:
{{{ (x+(-1))^2/4 -4(y+1)^2/4 = 1}}}
The 4's cancel in the "y fraction" (which is nice because we didn't want a coefficient anyway):
{{{ (x+(-1))^2/4 -(y+1)^2 = 1}}}
The first denominator is obviously a perfect square. But there is no denominator at all for the y term. But its easy to get one:
{{{ (x+(-1))^2/4 -(y+1)^2/1 = 1}}}
and a 1 is also a prefect square.
{{{ (x+(-1))^2/2^2 -(y+1)^2/1^2 = 1}}}
And last of all, rewrite the completed squares as subtractions:
{{{ (x-1)^2/2^2 -(y-(-1))^2/1^2 = 1}}}<br>
From this we can see that the center is (1, -1), the "a" is 2 and the "b" is 1.