SOLUTION: Put in standard form and identify the conic 1) y^2-x^2+6x-4y=6 2)x^2+y^2+14y=-13 3) 4x^2+9y^2+16x-54y=-61 4) x^2+4y^2-2x-15=0 I'm very confused any help would be very appre

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Put in standard form and identify the conic 1) y^2-x^2+6x-4y=6 2)x^2+y^2+14y=-13 3) 4x^2+9y^2+16x-54y=-61 4) x^2+4y^2-2x-15=0 I'm very confused any help would be very appre      Log On


   



Question 871001: Put in standard form and identify the conic
1) y^2-x^2+6x-4y=6
2)x^2+y^2+14y=-13
3) 4x^2+9y^2+16x-54y=-61
4) x^2+4y^2-2x-15=0
I'm very confused any help would be very appreciated

Found 2 solutions by richwmiller, lwsshak3:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
You are confused because the rules clearly state that you should submit only one problem at a time and no similar problems.
(y-2)^2-(x-3)^2 = 1 hyperbola for #1

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Put in standard form and identify the conic
1)y^2-x^2+6x-4y=6
y^2-4y-x^2+6x=6
complete the square
(y^2-4y+4)-(x^2-6x+9)=6+4-9
(y-2)^2-(x-3)^2=1
This is an equation of a hyperbola with vertical transverse axis and center at (3,2)
Its standard form of equation: %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, (h,k)=coordinates of center.
..
2)x^2+y^2+14y=-13
complete the square
x^2+(y^2+14y+49)=-13+49
x^2+(y+7)^2=36
This is an equation of a circle with center at (0,-7) and radius=6
Its standard form of equation: %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2,(h,k)=coordinates of center, r=radius
..
3) 4x^2+9y^2+16x-54y=-61
4x^2+16x+9y^2-54y=-61
complete the square
4(x^2+4x+4)+9(y^2-6y+9)=-61+16=81
4(x+2)^2+9(y-3)^2=36
(x+2)^2/9+(y-3)^2/4=1
This is an equation of an ellipse with horizontal major axis and center at (-2,3)
Its standard form of equation: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1,a>b,(h,k)=coordinates of center.
..
4)x^2+4y^2-2x-15=0
x^2-2x+4y^2-15=0
complete the square
(x^2-2x+1)+4y^2=15+1
(x-1)^2+4y^2=16
(x-1)^2/16+y^2/4=1
This is an equation of an ellipse with horizontal major axis and center at (1,0)
Its standard form of equation: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1,a>b,(h,k)=coordinates of center.