Convert the following equations of hyperbolas into their standard form.
x^2-y^2-4x-4y=4
To convert the equation into the standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2-4x-4y=4
\left(x^2-4x\right)-\left(y^2+4y\right)=4
\left[x^2-4x + \left(\dfrac{4}{2}\right)^2- \left(\dfrac{4}{2}\right)^2\right] -\left[y^2+4y +\left(\dfrac{4}{2}\right)^2-\left(\dfrac{4}{2}\right)^2\right]=4
\left[x^2-4x +4-4\right] -\left[y^2+4y +4-4\right]=4
\left(x^2-4x + 4\right)- 4 -\left(y^2+4y +4\right)+ 4=4
\left(x-2\right)^2-\left(y-2\right)^2=4
\dfrac{\left(x - 2\right)^2}4 - \dfrac{\left(y - 2\right)^2} 4 = 1
The standard form of the equation is:
\dfrac{\left(x - 2\right)^2}4 - \dfrac{\left(y - 2\right)^2} 4 = 1
x^2-y^2-6x+2y=1
To convert the equation into the standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2-6x+2y=1
\left(x^2-6x\right)-\left(y^2-2y\right)=1
\left[x^2-6x + \left(\dfrac{6}{2}\right)^2- \left(\dfrac{6}{2}\right)^2\right] -\left[y^2-2y +\left(\dfrac{2}{2}\right)^2-\left(\dfrac{2}{2}\right)^2\right]=1
\left[x^2-6x +9-9\right] -\left[y^2-2y +1-1\right]=1
\left(x^2-6x + 9\right)- 9 -\left(y^2-2y +1\right)+1=1
\left(x-3\right)^2-\left(y-1\right)^2=9
\dfrac{\left(x - 3\right)^2}9 - \dfrac{\left(y - 1\right)^2} 9 = 1
The standard form of the equation is:
\dfrac{\left(x - 3\right)^2}9 - \dfrac{\left(y - 1\right)^2} 9 = 1
x^2-y^2+2x+4y=13
To convert the equation into the standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2+2x+4y=13
\left(x^2+2x\right)-\left(y^2-4y\right)=13
\left[x^2+2x + \left(\dfrac{2}{2}\right)^2- \left(\dfrac{2}{2}\right)^2\right] -\left[y^2-4y +\left(\dfrac{4}{2}\right)^2-\left(\dfrac{4}{2}\right)^2\right]=13
\left[x^2+2x+1-1\right] -\left[y^2-4y +4-4\right]=13
\left(x^2+2x + 1\right)- 1 -\left(y^2-4y +4\right)+ 4=13
\left(x+1\right)^2-\left(y-2\right)^2=10
\dfrac{\left(x +1\right)^2}{10} - \dfrac{\left(y - 2\right)^2}{10} = 1
The standard form of the equation is:
\dfrac{\left(x +1\right)^2}{10} - \dfrac{\left(y - 2\right)^2}{10} = 1
x^2-y^2-4x-2y=9
To convert the equation into the standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2-4x-2y=9
\left(x^2-4x\right)-\left(y^2+2y\right)=9
\left[x^2-4x + \left(\dfrac{4}{2}\right)^2- \left(\dfrac{4}{2}\right)^2\right] -\left[y^2+2y +\left(\dfrac{2}{2}\right)^2-\left(\dfrac{2}{2}\right)^2\right]=9
\left[x^2-4x +4-4\right] -\left[y^2+2y +1-1\right]=9
\left(x^2-4x + 4\right)- 4 -\left(y^2+2y +1\right)+ 1=9
\left(x-2\right)^2-\left(y+1\right)^2=12
\dfrac{\left(x - 2\right)^2}{12} - \dfrac{\left(y+1\right)^2} {12} = 1
The standard form of the equation is:
\dfrac{\left(x - 2\right)^2}{12} - \dfrac{\left(y+1\right)^2} {12} = 1
x^2-y^2+6x+4y=20
To convert the equation into the standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2+6x+4y=20
\left(x^2+6x\right)-\left(y^2-4y\right)=20
\left[x^2+6x + \left(\dfrac{6}{2}\right)^2- \left(\dfrac{6}{2}\right)^2\right] -\left[y^2-4y +\left(\dfrac{4}{2}\right)^2-\left(\dfrac{4}{2}\right)^2\right]=20
\left[x^2+6x +9-9\right] -\left[y^2-4y +4-4\right]=20
\left(x^2+6x +9\right)-9 -\left(y^2-4y +4\right)+ 4=20
\left(x+3\right)^2-\left(y-2\right)^2=25
\dfrac{\left(x +3\right)^2}{25} - \dfrac{\left(y - 2\right)^2} {25} = 1
The standard form of the equation is:
\dfrac{\left(x +3\right)^2}{25} - \dfrac{\left(y - 2\right)^2} {25} = 1
x^2-y^2-8x-2y=-10
To convert the equation into standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 :
x^2-y^2-8x-2y=-10
\left(x^2-8x\right)-\left(y^2+2y\right)=-10
\left[x^2-8x + \left(\dfrac{8}{2}\right)^2- \left(\dfrac{8}{2}\right)^2\right] -\left[y^2+2y +\left(\dfrac{2}{2}\right)^2-\left(\dfrac{2}{2}\right)^2\right]=-10
\left[x^2-8x +16-16\right] -\left[y^2+2y +1-1\right]=-10
\left(x^2-8x + 16\right)-16 -\left(y^2+2y +1\right)+1=-10
\left(x-4\right)^2-\left(y+1\right)^2=5
\dfrac{\left(x - 4\right)^2}5 - \dfrac{\left(y + 1\right)^2}5 = 1
The standard form of the equation is:
\dfrac{\left(x - 4\right)^2}5 - \dfrac{\left(y + 1\right)^2}5 = 1
x^2-y^2-10x-4y=-14
To convert the equation into standard form, complete the square. Assume that an expression is given:
aX^2+ bX
To complete the square, add and subtract \left(\dfrac{b}{2a}\right)^2 e:
x^2-y^2-10x-4y=-14
\left(x^2-10x\right)-\left(y^2+4y\right)=-14
\left[x^2-10x + \left(\dfrac{10}{2}\right)^2- \left(\dfrac{10}{2}\right)^2\right] -\left[y^2+4y +\left(\dfrac{4}{2}\right)^2-\left(\dfrac{4}{2}\right)^2\right]=-14
\left[x^2-10x +25-25\right] -\left[y^2+4y +4-4\right]=-14
\left(x^2-10x +25\right)- 25 -\left(y^2+4y +4\right)+ 4=-14
\left(x-5\right)^2-\left(y+2\right)^2=7
\dfrac{\left(x - 5\right)^2}7 - \dfrac{\left(y+ 2\right)^2} 7 = 1
The standard form of the equation is:
\dfrac{\left(x - 5\right)^2}7 - \dfrac{\left(y+ 2\right)^2} 7 = 1