Find the equation of ellipse that satifies the given conditions:
(a) Vertices `(pm 6, 0) , ` foci `(pm 4, 0 ) `
(b) Ends of major axis `(0 , pm sqrt(5))` : ends of minor axis `(pm 1, 0)`

Answer

To find the equations of the ellipse that satisfies the given conditions, we will solve each part step by step.

### Part (a)

**Given:**
- Vertices: \( (\pm 6, 0) \)
- Foci: \( (\pm 4, 0) \)

**Step 1: Identify the values of \( a \) and \( c \)**

The vertices of the ellipse are located at \( (\pm a, 0) \). Here, the vertices are \( (\pm 6, 0) \), so:
\[ a = 6 \]

The foci of the ellipse are located at \( (\pm c, 0) \). Here, the foci are \( (\pm 4, 0) \), so:
\[ c = 4 \]

**Step 2: Use the relationship between \( a \), \( b \), and \( c \)**

For an ellipse, the relationship is given by:
\[ c^2 = a^2 - b^2 \]

Substituting the known values:
\[ 4^2 = 6^2 - b^2 \]
\[ 16 = 36 - b^2 \]
\[ b^2 = 36 - 16 = 20 \]

**Step 3: Write the equation of the ellipse**

Since the major axis is along the x-axis, the equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = 36 \) and \( b^2 = 20 \):
\[ \frac{x^2}{36} + \frac{y^2}{20} = 1 \]

### Solution for Part (a):
The equation of the ellipse is:
\[ \frac{x^2}{36} + \frac{y^2}{20} = 1 \]

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### Part (b)

**Given:**
- Ends of major axis: \( (0, \pm \sqrt{5}) \)
- Ends of minor axis: \( (\pm 1, 0) \)

**Step 1: Identify the values of \( a \) and \( b \)**

The ends of the major axis are located at \( (0, \pm a) \). Here, the ends are \( (0, \pm \sqrt{5}) \), so:
\[ a = \sqrt{5} \]
Thus,
\[ a^2 = 5 \]

The ends of the minor axis are located at \( (\pm b, 0) \). Here, the ends are \( (\pm 1, 0) \), so:
\[ b = 1 \]
Thus,
\[ b^2 = 1 \]

**Step 2: Write the equation of the ellipse**

Since the major axis is along the y-axis, the equation of the ellipse is:
\[ \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \]
Substituting \( a^2 = 5 \) and \( b^2 = 1 \):
\[ \frac{y^2}{5} + \frac{x^2}{1} = 1 \]

### Solution for Part (b):
The equation of the ellipse is:
\[ \frac{y^2}{5} + x^2 = 1 \]

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