Find the equation of the ellipse satisfying the following conditions:
(i) Vertices at (`pm 13, 0`), foci at `(pm 5,0)`
(ii) Foci at `(pm 3, 0 )` passing through (4,1 )

Answer

To find the equations of the ellipses based on the given conditions, we will solve each part step by step.

### Part (i): Vertices at (±13, 0) and foci at (±5, 0)

1. **Identify the values of a and c:**
- The vertices of the ellipse are at (±13, 0). Therefore, the distance from the center to a vertex (a) is 13.
- The foci are at (±5, 0). Therefore, the distance from the center to a focus (c) is 5.

\[ a = 13, \quad c = 5 \]

2. **Calculate b using the relationship \(c^2 = a^2 - b^2\):**
- We know \(c^2 = a^2 - b^2\).
- Substitute the values of a and c:

\[ 5^2 = 13^2 - b^2 \]
\[ 25 = 169 - b^2 \]
\[ b^2 = 169 - 25 = 144 \]
\[ b = 12 \]

3. **Write the standard form of the ellipse equation:**
- Since the major axis is along the x-axis, the standard form of the ellipse is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Substituting the values of \(a^2\) and \(b^2\):

\[ \frac{x^2}{169} + \frac{y^2}{144} = 1 \]

### Part (ii): Foci at (±3, 0) and passing through (4, 1)

1. **Identify the values of c:**
- The foci are at (±3, 0), so \(c = 3\).

2. **Let a be the distance from the center to a vertex.**
- We do not know \(a\) yet, but we will find it using the point (4, 1) that lies on the ellipse.

3. **Use the relationship \(c^2 = a^2 - b^2\):**
- We will express \(b^2\) in terms of \(a\):

\[ 3^2 = a^2 - b^2 \implies b^2 = a^2 - 9 \]

4. **Substitute the point (4, 1) into the ellipse equation:**
- The standard form of the ellipse is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Substituting (4, 1):

\[ \frac{4^2}{a^2} + \frac{1^2}{b^2} = 1 \]
\[ \frac{16}{a^2} + \frac{1}{a^2 - 9} = 1 \]

5. **Multiply through by \(a^2(a^2 - 9)\) to eliminate the denominators:**

\[ 16(a^2 - 9) + a^2 = a^2(a^2 - 9) \]
\[ 16a^2 - 144 + a^2 = a^4 - 9a^2 \]
\[ a^4 - 9a^2 - 17a^2 + 144 = 0 \]
\[ a^4 - 26a^2 + 144 = 0 \]

6. **Let \(u = a^2\), then solve the quadratic equation:**
\[ u^2 - 26u + 144 = 0 \]

Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ u = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 144}}{2 \cdot 1} \]
\[ u = \frac{26 \pm \sqrt{676 - 576}}{2} \]
\[ u = \frac{26 \pm \sqrt{100}}{2} \]
\[ u = \frac{26 \pm 10}{2} \]
\[ u = 18 \quad \text{or} \quad u = 8 \]

Thus, \(a^2 = 18\) or \(a^2 = 8\).

7. **Calculate b for both cases:**
- For \(a^2 = 18\):
\[ b^2 = 18 - 9 = 9 \quad \Rightarrow \quad b = 3 \]
- For \(a^2 = 8\):
\[ b^2 = 8 - 9 \quad \text{(not possible since } b^2 < 0\text{)} \]

8. **Write the equation of the ellipse:**
- Using \(a^2 = 18\) and \(b^2 = 9\):

\[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \]

### Final Answers:
1. For the first ellipse: \(\frac{x^2}{169} + \frac{y^2}{144} = 1\)
2. For the second ellipse: \(\frac{x^2}{18} + \frac{y^2}{9} = 1\)