The given equation is x² + 4y – 2y². To identify the type of conic section, we can rearrange the equation into the standard form.
First, let’s rewrite it:
2y² – 4y – x² = 0.
This can be expressed in the standard form of a conic section. To do this, we can complete the square for the y-terms:
2(y² – 2y) = x².
Completing the square, we add and subtract 1 inside the parentheses:
2((y – 1)² – 1) = x²
Which simplifies to:
2(y – 1)² – 2 = x²
Rearranging gives us:
2(y – 1)² = x² + 2.
Dividing everything by 2 yields:
(y – 1)²/1 – x²/2 = 1.
This is now in the standard form of a hyperbola:
(y – k)²/a² – (x – h)²/b² = 1,
Where (h, k) is the center. Here, we identify that:
- (h, k) = (0, 1)
- a² = 1 and b² = 2
The vertices of a hyperbola given in this standard form are found at (h, k ± a):
- Vertices: (0, 1 + 1) = (0, 2) and (0, 1 – 1) = (0, 0)
To find the foci, we use the formula c = √(a² + b²):
- c = √(1 + 2) = √3
The foci are at (h, k ± c):
- Foci: (0, 1 + √3) and (0, 1 – √3)
In conclusion, the equation x² + 4y – 2y² describes a hyperbola with vertices at (0, 2) and (0, 0), and foci at (0, 1 + √3) and (0, 1 – √3).