28. Center: $$ (3,2) ; a=3 c $$; foci: $$ (1,2),(5,2) $$ 29. Vertices: $$ (0,2),(4,2) $$; endpoints of the minor axis: $$ (2,3),(2,1) $$ 30. Vertices: $$ (5,0),(5,12) $$; endpoints of the minor axis: $$ (1,6),(9,6) $$

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To solve the problems related to the given ellipses, we will extract the known conditions and derive the equations of the ellipses step by step.

### Problem 28
**Given:**
- Center: $$ (3,2) $$
- Foci: $$ (1,2) $$ and $$ (5,2) $$
- $$ a = 3c $$

**Step 1: Identify the orientation of the ellipse.**
Since the foci have the same $$ y $$-coordinate, the ellipse is horizontal.

**Step 2: Calculate the distance between the foci.**
The distance between the foci is given by:
$$
d = 5 - 1 = 4
$$
Thus, $$ 2c = 4 $$ implies $$ c = 2 $$.

**Step 3: Calculate $$ a $$.**
Given $$ a = 3c $$:
$$
a = 3 \times 2 = 6
$$

**Step 4: Calculate $$ b $$ using the relationship $$ c^2 = a^2 - b^2 $$.**
$$
c^2 = 2^2 = 4
$$
$$
a^2 = 6^2 = 36
$$
Thus,
$$
4 = 36 - b^2 \implies b^2 = 36 - 4 = 32
$$

**Step 5: Write the equation of the ellipse.**
The standard form of the ellipse is:
$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$
Substituting $$ h = 3 $$, $$ k = 2 $$, $$ a^2 = 36 $$, and $$ b^2 = 32 $$:
$$
\frac{(x - 3)^2}{36} + \frac{(y - 2)^2}{32} = 1
$$

### Problem 29
**Given:**
- Vertices: $$ (0,2) $$ and $$ (4,2) $$
- Endpoints of the minor axis: $$ (2,3) $$ and $$ (2,1) $$

**Step 1: Identify the orientation of the ellipse.**
Since the vertices have the same $$ y $$-coordinate, the ellipse is horizontal.

**Step 2: Calculate $$ a $$.**
The distance between the vertices is:
$$
d = 4 - 0 = 4 \implies 2a = 4 \implies a = 2
$$

**Step 3: Calculate $$ b $$.**
The distance between the endpoints of the minor axis is:
$$
d = 3 - 1 = 2 \implies 2b = 2 \implies b = 1
$$

**Step 4: Calculate the center.**
The center $$ (h, k) $$ is the midpoint of the vertices:
$$
h = \frac{0 + 4}{2} = 2, \quad k = 2
$$

**Step 5: Write the equation of the ellipse.**
$$
\frac{(x - 2)^2}{2^2} + \frac{(y - 2)^2}{1^2} = 1
$$
This simplifies to:
$$
\frac{(x - 2)^2}{4} + (y - 2)^2 = 1
$$

### Problem 30
**Given:**
- Vertices: $$ (5,0) $$ and $$ (5,12) $$
- Endpoints of the minor axis: $$ (1,6) $$ and $$ (9,6) $$

**Step 1: Identify the orientation of the ellipse.**
Since the vertices have the same $$ x $$-coordinate, the ellipse is vertical.

**Step 2: Calculate $$ a $$.**
The distance between the vertices is:
$$
d = 12 - 0 = 12 \implies 2a = 12 \implies a = 6
$$

**Step 3: Calculate $$ b $$.**
The distance between the endpoints of the minor axis is:
$$
d = 9 - 1 = 8 \implies 2b = 8 \implies b = 4
$$

**Step 4: Calculate the center.**
The center $$ (h, k) $$ is the midpoint of the vertices:
$$
h = 5, \quad k = \frac{0 + 12}{2} = 6
$$

**Step 5: Write the equation of the ellipse.**
$$
\frac{(x - 5)^2}{4^2} + \frac{(y - 6)^2}{6^2} = 1
$$
This simplifies to:
$$
\frac{(x - 5)^2}{16} + \frac{(y - 6)^2}{36} = 1
$$

### Summary of Results
1. **Problem 28:** $$\frac{(x - 3)^2}{36} + \frac{(y - 2)^2}{32} = 1$$
2. **Problem 29:** $$\frac{(x - 2)^2}{4} + (y - 2)^2 = 1$$
3. **Problem 30:** $$\frac{(x - 5)^2}{16} + \frac{(y - 6)^2}{36} = 1$$
The equations of the ellipses are as follows: 1. **Problem 28:** $$\frac{(x - 3)^2}{36} + \frac{(y - 2)^2}{32} = 1$$ 2. **Problem 29:** $$\frac{(x - 2)^2}{4} + (y - 2)^2 = 1$$ 3. **Problem 30:** $$\frac{(x - 5)^2}{16} + \frac{(y - 6)^2}{36} = 1$$

28. Center: \( (3,2) ; a=3 c \); foci: \( (1,2),(5,2) \) 29. Vertices: \( (0,2),(4,2) \); endpoints of the minor axis: \( (2,3),(2,1) \) 30. Vertices: \( (5,0),(5,12) \); endpoints of the minor axis: \( (1,6),(9,6) \)

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