document.write( "Question 612903: Please help me find the standard form of the equation : \"+x%5E2-2x-4y%5E2-8y-7=0\". By completing the square. And state the type of conic.
\n" ); document.write( "Please show all steps so I can understand how to get to the answer. \n" ); document.write( "
Algebra.Com's Answer #385795 by jsmallt9(3758)\"\" \"About 
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\"+x%5E2-2x-4y%5E2-8y-7=0\"
\n" ); document.write( "Because the squared terms have opposite signs, this equation is the equation of a hyperbola. The general standard form for a hyperbola is:
\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=+1\"
\n" ); document.write( "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:
  1. \"Move\" the constant term to one side of the equation (with all the variable terms on the other side).
  2. Rearrange the order of the terms on the variable side so that the x terms are together and the y terms are together.
  3. 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!
  4. For each variable, find half of the first power term.
  5. For each variable, find the square of the half you found in the previous step.
  6. 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).
  7. And last, rewrite as squares of binomials. The general form is:
    \n" ); document.write( "\"(x + the half from step 4) squared\" and \"(y + the half from step 4) squared\"
Let's see this in action:
\n" ); document.write( "1. Move the constant term.
\n" ); document.write( "Adding 7 to each side...
\n" ); document.write( "\"+x%5E2-2x-4y%5E2-8y=7\"
\n" ); document.write( "2. Rearrange the terms...
\n" ); document.write( "Your x terms and y terms are already together.
\n" ); document.write( "3. Factor out coefficients of the squared terms.
\n" ); document.write( "Your \"x%5E2\" has a coefficient of 1 so we can skip this. Your \"y%5E2\" term has a coefficient of -4:
\n" ); document.write( "\"+x%5E2-2x-4%28y%5E2%2B2y%29=7\"
\n" ); document.write( "Note how factoring out the -4 changes the sign of the +8y!
\n" ); document.write( "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.
\n" ); document.write( "5. Find the squares of the halves.
\n" ); document.write( "Both -1 squared and 1 squared is 1.
\n" ); document.write( "6. Add the squares to each side.
\n" ); document.write( "This is the hardest part to get right. This is how the left side should look;
\n" ); document.write( "\"+x%5E2-2x%2B+1+-4%28y%5E2%2B2y%2B+1%29\"
\n" ); document.write( "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!
\n" ); document.write( "\"+x%5E2-2x%2B+1+-4%28y%5E2%2B2y%2B+1%29+=+7+%2B+1+%2B+%28-4%29\"
\n" ); document.write( "Simplifying the right side we get:
\n" ); document.write( "\"+x%5E2-2x%2B+1+-4%28y%5E2%2B2y%2B+1%29+=+4\"
\n" ); document.write( "7. Rewrite as squares of binomials, using the \"halves\" from step 4:
\n" ); document.write( "\"+%28x%2B%28-1%29%29%5E2+-4%28y%2B1%29%5E2+=+4\"

\n" ); document.write( "After all that we have completed the squares. For the standard form we need:
  1. A 1 on the right side
  2. No coefficients in front of the completed squares
  3. Perfect square denominators under the completed squares.
  4. Rewrite the completed squares as subtractions.

\n" ); document.write( "To get the 1 on the right, divide both sides by the 4 that is there:
\n" ); document.write( "\"+%28x%2B%28-1%29%29%5E2%2F4+-4%28y%2B1%29%5E2%2F4+=+1\"
\n" ); document.write( "The 4's cancel in the \"y fraction\" (which is nice because we didn't want a coefficient anyway):
\n" ); document.write( "\"+%28x%2B%28-1%29%29%5E2%2F4+-%28y%2B1%29%5E2+=+1\"
\n" ); document.write( "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:
\n" ); document.write( "\"+%28x%2B%28-1%29%29%5E2%2F4+-%28y%2B1%29%5E2%2F1+=+1\"
\n" ); document.write( "and a 1 is also a prefect square.
\n" ); document.write( "\"+%28x%2B%28-1%29%29%5E2%2F2%5E2+-%28y%2B1%29%5E2%2F1%5E2+=+1\"
\n" ); document.write( "And last of all, rewrite the completed squares as subtractions:
\n" ); document.write( "\"+%28x-1%29%5E2%2F2%5E2+-%28y-%28-1%29%29%5E2%2F1%5E2+=+1\"

\n" ); document.write( "From this we can see that the center is (1, -1), the \"a\" is 2 and the \"b\" is 1. \n" ); document.write( "
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