SOLUTION: this is from the shifted conics sections, you have to identify whether it is a parabola, ellipse or hyperbola, and solve and graph the problem: 2x^2 + y^2 = 2y + 1

2x² + y² = 2y + 1

We place it in the form

Ax² + Bxy + Cy² + Dx + Ey + F = 0

2x² + 0xy + 1y² + 0x - 2y - 1 = 0

A = 2
B = 0
C = 1
D = 0
E = -2
F = -1

The rules are:

1. if the discriminant, B² − 4AC < 0, the equation represents an ellipse; 
2. if A = C and B = 0, the equation represents a circle; 
3. if B2 − 4AC = 0, the equation represents a parabola; 
4. if B2 − 4AC > 0, the equation represents a hyperbola; 
5. if we also have A + C = 0, the equation represents a rectangular hyperbola. 

The discriminant is B²-4AC = 0²-4(2)(1) = -8, so it represents an ellipse.

Since it has no xy term, we get it into one of these forms

 or 

Either way we begin by getting the constant on the right side:

2x² + y² - 2y = 1

We complete the square on y by adding 1 to both sides:

2x² + y² - 2y + 1 = 1 + 1

We factor the last three terms on the left as a perfect square:

2x² + (y - 1)² = 2

We divide through by 2 to get 1 on the right side:





We write x as (x-0)



Write the first term over 1



That is the form 



because a is always greater than b in an ellipse.

It has center (h,k) = (0,1), semi-major axis = a = 1,
and semi-minor axis = b = 

It has its major axis vertical because a² is under the term in y

We plot the center:



We draw the complete major axis , about 1.4 units up and down
from the center (in purple):



We draw the complete minor axis  unit right and left
from the center (in green):
 



We sketch in the ellipse:




Edwin