SOLUTION: please help me solve this question: y^2+2y=x^2+x is this equation a parabola, circle, ellipse, or hyperbola? how do you know? thnks

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Question 315531: please help me solve this question:
y^2+2y=x^2+x
is this equation a parabola, circle, ellipse, or hyperbola? how do you know?
thnks

Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
please help me solve this question:
y2 + 2y = x2 + x
is this equation a parabola, circle, ellipse, or hyperbola? how do you know?
thnks

Get it in the standard conic form:

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

then calculate the discriminant, which is B2 - 4AC

1. If the discriminant is positive, the equation represents a hyperbola.

2. If the discriminant is zero, the equation represents a parabola.

3. If the discriminant is negative then it is either an ellipse or a circle.
   If it's a circle, then B=0 and A=C, otherwise it's an ellipse.

So let's get

y2 + 2y = x2 + x

in the standard conic form 

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

Get 0 on the right by subtracting the right side from both sides:

y2 + 2y - x2 - x = 0

Now lets rearrange the terms in the order as they appear in the standard
conic form:

 Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

-1x2 + 0xy + 1y2 - 1x + 2y + 0 = 0
 
So A=-1, B=0, C=1, D=-1, E=2, F=0

B2 - 4AC = (0)^2 - 4(-1)(1) = 0 - (-4) = 0 + 4 = 4

That's positive, so the equation represents a hyperbola.

-----------------------------------

Edwin

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
y^2+2y=x^2+x
is this equation a parabola, circle, ellipse, or hyperbola? how do you know?
It is a hypobola because the x^2 and y^2 have opposite signs when
arranged on the same side of the equation.
=================================================
Rearrange:
x^2 + x - (y^2+2y) = 0
Complete the square:
(x^2 + x + (1/2)^2) - (y^2+2y+1) = (1/2)^2 - 1
(x+1/2)^2 - (y+1)^2 = -3/4
----
(y+1)^2 - (x+(1/2))^2 = (3/4)
===============================
Cheers,
Stan H.
===============================

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