SOLUTION: PART 1: The graph of the equation y^2-x^2+2x=2 is: PART 2: The graph of the equation y-x^2=2x a. parabola b. elipse c. circle d. hyperbola e. none of these Part 1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: PART 1: The graph of the equation y^2-x^2+2x=2 is: PART 2: The graph of the equation y-x^2=2x a. parabola b. elipse c. circle d. hyperbola e. none of these Part 1       Log On


   

Question 692383: PART 1: The graph of the equation y^2-x^2+2x=2 is:
PART 2: The graph of the equation y-x^2=2x

a. parabola
b. elipse
c. circle
d. hyperbola
e. none of these


Part 1 Solve:
y^2-x^2+2x=2
???????????
Part 2 Solve:
y-x^2=2x
y=x^2+2x
(parabola)


That's all I got, I don't know how I would solve part 1 to figure out what kind it is and I'm pretty sure part 2 is right. If anyone can solve part 1 it would help a lot
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
PART 1: The graph of the equation y^2-x^2+2x=2 is:
complete the square:
y^2-x^2+2x=2
y^2-(x^2-2x+1)=2-1
y^2-(x-1)^2=1
This is an equation of a hyperbola with vertical transverse axis and center at (1,0)
Its standard form: %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center
..
PART 2: The graph of the equation y-x^2=2x
complete the square:
y-x^2=2x
y-x^2-2x=0
y-(x^2+2x+1)=0-1
y-(x+1)^2=-1
(x+1)^2=y+1
This is an equation of a parabola that opens upwards.
Its standard form: %28x-h%29%5E2=4p%28y-k%29, (h,k)=(x,y) coordinates of the vertex
note: Often the best way to see what conic the given equation is, write it in standard form by completing the square.