Find the equation of the ellipse referred to its axes as the axes of co-ordinates :
(i) whose major axis = 6 and minor axis = 4
(ii) Which passes through the point (-3, 1) and has eccentricity = `sqrt((2)/(5))`
(iii) whose foci are (2,0) , (-2,0) and latus -rectum is 6.

Answer

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

### Part (i)
**Given:**
- Major axis = 6
- Minor axis = 4

**Solution:**
1. The lengths of the major and minor axes are given as 2a and 2b respectively.
- Therefore, \( 2a = 6 \) implies \( a = 3 \).
- And \( 2b = 4 \) implies \( b = 2 \).

2. The standard equation of an ellipse with a horizontal major axis is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

3. Substituting the values of \( a \) and \( b \):
\[ \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

4. To express it in a different form, multiply through by 36 (the least common multiple of the denominators):
\[ 4x^2 + 9y^2 = 36 \]

**Final Equation:**
\[ 4x^2 + 9y^2 = 36 \]