To find the center, vertices, and foci of the ellipse defined by the equation 4x² + 6y² = 24, we first need to rewrite it in standard form.
1. **Rewrite the Equation:**
We start with the given equation:
4x² + 6y² = 24
Next, we divide each term by 24 to normalize the equation:
\( \frac{4x²}{24} + \frac{6y²}{24} = 1 \
Simplifying further gives us:
\( \frac{x²}{6} + \frac{y²}{4} = 1 \
2. **Identify the Center:**
The standard form of the ellipse is given by:
\( \frac{(x-h)²}{a²} + \frac{(y-k)²}{b²} = 1 \,
where (h, k) is the center. From our equation:
We can identify:
- h = 0
- k = 0
Therefore, the center of the ellipse is at the point (0, 0).
3. **Determine the Vertices:**
From our rewritten equation, we can see:
a² = 6 and b² = 4, thus:
a = √6 and b = 2
Since the larger denominator is associated with the x² term, this indicates that the major axis is horizontal. The vertices are located at:
- (h ± a, k) = (0 ± √6, 0) -> Vertices: ((√6, 0), (-√6, 0))
4. **Locate the Foci:**
The distance of the foci from the center is given by:
c = √(a² – b²).
Substituting our values gives:
c = √(6 – 4) = √2
The foci will be positioned along the major axis, thus:
- (h ± c, k) = (0 ± √2, 0) -> Foci: ((√2, 0), (-√2, 0))
5. **Summary:**
To summarize:
- Center: (0, 0)
- Vertices: (√6, 0) and (-√6, 0)
- Foci: (√2, 0) and (-√2, 0)
This process outlines how to derive the center, vertices, and foci of the referred ellipse effectively.