16. Horizontal major axis; passes through the points $$ (5,0) $$ and $$ (0,2) $$ 17. Vertices: $$ ( \pm 6,0) $$; passes through the point $$ (4,1) $$ 18. Vertices: $$ (0, \pm 8) $$; passes through the point $$ (3,4) $$

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To solve the problems related to the ellipses, we will use the standard forms of the equations of ellipses and the given conditions.

### Problem 16: Horizontal Major Axis
1. **Known Conditions:**
   - Major axis is horizontal.
   - Passes through points $$ (5,0) $$ and $$ (0,2) $$.

2. **Standard Form:**
   The standard form of an ellipse with a horizontal major axis is:
   $$
   \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1
   $$
   where $$ (h, k) $$ is the center, $$ a $$ is the semi-major axis, and $$ b $$ is the semi-minor axis.

3. **Finding the Center:**
   The center $$ (h, k) $$ is the midpoint of the points $$ (5,0) $$ and $$ (0,2) $$:
   $$
   h = \frac{5 + 0}{2} = 2.5, \quad k = \frac{0 + 2}{2} = 1
   $$
   So, the center is $$ (2.5, 1) $$.

4. **Finding $$ a $$ and $$ b $$:**
   The distance from the center to the vertices along the major axis (horizontal) is $$ a $$. The distance from the center to the minor axis is $$ b $$.

- The distance from the center $$ (2.5, 1) $$ to $$ (5, 0) $$:
   $$
   a = \sqrt{(5 - 2.5)^2 + (0 - 1)^2} = \sqrt{(2.5)^2 + (-1)^2} = \sqrt{6.25 + 1} = \sqrt{7.25}
   $$

- The distance from the center $$ (2.5, 1) $$ to $$ (0, 2) $$:
   $$
   b = \sqrt{(0 - 2.5)^2 + (2 - 1)^2} = \sqrt{(-2.5)^2 + (1)^2} = \sqrt{6.25 + 1} = \sqrt{7.25}
   $$

5. **Equation of the Ellipse:**
   The equation becomes:
   $$
   \frac{(x - 2.5)^2}{7.25} + \frac{(y - 1)^2}{7.25} = 1
   $$

### Problem 17: Vertices $$ ( \pm 6,0) $$
1. **Known Conditions:**
   - Vertices: $$ ( \pm 6,0) $$.
   - Passes through the point $$ (4,1) $$.

2. **Standard Form:**
   The standard form is the same as above. Here, $$ a = 6 $$ (the distance from the center to the vertices).

3. **Finding the Center:**
   The center is at $$ (0, 0) $$.

4. **Finding $$ b $$:**
   We need to find $$ b $$ using the point $$ (4,1) $$:
   $$
   \frac{(4 - 0)^2}{6^2} + \frac{(1 - 0)^2}{b^2} = 1
   $$
   Simplifying:
   $$
   \frac{16}{36} + \frac{1}{b^2} = 1 \implies \frac{4}{9} + \frac{1}{b^2} = 1
   $$
   $$
   \frac{1}{b^2} = 1 - \frac{4}{9} = \frac{5}{9} \implies b^2 = \frac{9}{5}
   $$

5. **Equation of the Ellipse:**
   The equation becomes:
   $$
   \frac{x^2}{36} + \frac{y^2}{\frac{9}{5}} = 1
   $$

### Problem 18: Vertices $$ (0, \pm 8) $$
1. **Known Conditions:**
   - Vertices: $$ (0, \pm 8) $$.
   - Passes through the point $$ (3,4) $$.

2. **Standard Form:**
   The standard form is:
   $$
   \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1
   $$
   Here, $$ a = 8 $$.

3. **Finding the Center:**
   The center is at $$ (0, 0) $$.

4. **Finding $$ b $$:**
   Using the point $$ (3,4) $$:
   $$
   \frac{(3 - 0)^2}{b^2} + \frac{(4 - 0)^2}{8^2} = 1
   $$
   Simplifying:
   $$
   \frac{9}{b^2} + \frac{16}{64} = 1 \implies \frac{9}{b^2} + \frac{1}{4} = 1
   $$
   $$
   \frac{9}{b^2} = 1 - \frac{1}{4} = \frac{3}{4} \implies b^2 = \frac{36}{3} = 12
   $$

5. **Equation of the Ellipse:**
   The equation becomes:
   $$
   \frac{x^2}{12} + \frac{y^2}{64} = 1
   $$

### Summary of Results
1. **Problem 16:** 
   $$
   \frac{(x - 2.5)^2}{7.25} + \frac{(y - 1)^2}{7.25} = 1
   $$
2. **Problem 17:** 
   $$
   \frac{x^2}{36} + \frac{y^2}{\frac{9}{5}} = 1
   $$
3. **Problem 18:** 
   $$
   \frac{x^2}{12} + \frac{y^2}{64} = 1
   $$
The equations of the ellipses are: 1. **Problem 16:** $$ \frac{(x - 2.5)^2}{7.25} + \frac{(y - 1)^2}{7.25} = 1 $$ 2. **Problem 17:** $$ \frac{x^2}{36} + \frac{y^2}{1.8} = 1 $$ 3. **Problem 18:** $$ \frac{x^2}{12} + \frac{y^2}{64} = 1 $$

16. Horizontal major axis; passes through the points \( (5,0) \) and \( (0,2) \) 17. Vertices: \( ( \pm 6,0) \); passes through the point \( (4,1) \) 18. Vertices: \( (0, \pm 8) \); passes through the point \( (3,4) \)

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