To identify the type of conic section from the given equation x² – 4y + 2y² = 0, we first need to rearrange and rewrite it in a standard form.
1. **Rearranging the Equation**:
We start by rearranging the equation:
2y² – 4y + x² = 0
2. **Completing the Square**:
We will complete the square for the y terms:
2(y² – 2y) + x² = 0
Taking 2 out:
2[(y – 1)² – 1] + x² = 0
Expanding it gives us:
2(y – 1)² – 2 + x² = 0
Now, we can rearrange that to:
x² + 2(y – 1)² = 2
3. **Standard Form**:
Now, we can write this as:
\( \frac{x²}{2} + \frac{(y – 1)²}{1} = 1 \)
This is the standard form of an ellipse.
4. **Identifying Vertices and Foci**:
For an ellipse in the standard form: \( \frac{(x – h)²}{a²} + \frac{(y – k)²}{b²} = 1 \), with center \((h,k)\), the vertices can be found at:
- Vertices along the x-axis: \((h ± a, k)\)
- Vertices along the y-axis: \((h, k ± b)\)
Here, \(h = 0\), \(k = 1\), \(a = \sqrt{2}\), and \(b = 1\).
Thus, the vertices are at:
- Horizontal vertices: \((\pm \sqrt{2}, 1)\)
- Vertical vertices: \((0, 1 ± 1) = (0, 0) \text{ and } (0, 2)\)
5. **Foci Calculation**:
The distance to the foci from the center can be found using the formula \(c = \sqrt{a² – b²}\):
Where:
– \(a = \sqrt{2}\)
– \(b = 1\)
Calculating \(c\):
c = \sqrt{(\sqrt{2})² – 1²} = \sqrt{2 – 1} = \sqrt{1} = 1
So, the foci are located at:
- Foci along the x-axis: \((h ± c, k) = (0 ± 1, 1) = (1, 1) \text{ and } (-1, 1)\)
Conclusion:
The type of conic section represented by the equation x² – 4y + 2y² = 0 is an ellipse with:
- Vertices: (\(\sqrt{2}, 1\)), (-\(\sqrt{2}, 1\)), (0, 2), (0, 0)
- Foci: (1, 1) and (-1, 1)