Find the equation of the ellipse in the standard form if
(i) e = `1/2` and Passes through (2, 1)
(ii) it passes through the points (-2,2), (3, -1)

Answer

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

### Part (i): Given e = 1/2 and passes through (2, 1)

1. **Understanding the eccentricity (e)**:
The eccentricity of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Given \( e = \frac{1}{2} \), we can square both sides:
\[ \left(\frac{1}{2}\right)^2 = 1 - \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{1}{4} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Therefore, we can express \( b^2 \) in terms of \( a^2 \):
\[ b^2 = \frac{3}{4} a^2 \quad \text{(Equation 1)} \]

2. **Using the point (2, 1)**:
The standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the point (2, 1) into the equation:
\[ \frac{2^2}{a^2} + \frac{1^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{1}{b^2} = 1 \]

3. **Substituting \( b^2 \) from Equation 1**:
Replace \( b^2 \) in the equation:
\[ \frac{4}{a^2} + \frac{1}{\frac{3}{4} a^2} = 1 \]
This becomes:
\[ \frac{4}{a^2} + \frac{4}{3a^2} = 1 \]
Finding a common denominator:
\[ \frac{12 + 4}{3a^2} = 1 \implies \frac{16}{3a^2} = 1 \]
Thus:
\[ 3a^2 = 16 \implies a^2 = \frac{16}{3} \]

4. **Finding \( b^2 \)**:
Substitute \( a^2 \) back into Equation 1:
\[ b^2 = \frac{3}{4} \cdot \frac{16}{3} = 4 \]

5. **Final equation of the ellipse**:
Substitute \( a^2 \) and \( b^2 \) into the standard form:
\[ \frac{x^2}{\frac{16}{3}} + \frac{y^2}{4} = 1 \]
Multiplying through by 12 to eliminate the denominators:
\[ 3x^2 + 12y^2 = 16 \]

### Part (ii): Passes through points (-2, 2) and (3, -1)

1. **Setting up the equations**:
Using the standard form of the ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substitute the point (-2, 2):
\[ \frac{(-2)^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{4}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 1)} \]

Substitute the point (3, -1):
\[ \frac{3^2}{a^2} + \frac{(-1)^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 2)} \]

2. **Solving the equations**:
From Equation 1:
\[ 4b^2 + 4a^2 = a^2b^2 \implies 4b^2 = a^2b^2 - 4a^2 \]
Rearranging gives:
\[ 4b^2 = a^2b^2 - 4a^2 \implies 4b^2 + 4a^2 = a^2b^2 \]

From Equation 2:
\[ 9b^2 + a^2 = a^2b^2 \]

3. **Eliminating variables**:
Multiply Equation 1 by \( b^2 \) and Equation 2 by \( a^2 \):
\[ 4b^2 + 4a^2 = a^2b^2 \quad \text{(1)} \]
\[ 9b^2 + a^2 = a^2b^2 \quad \text{(2)} \]

Subtract (1) from (2):
\[ (9 - 4)b^2 + (1 - 4)a^2 = 0 \implies 5b^2 - 3a^2 = 0 \implies b^2 = \frac{3}{5}a^2 \]

4. **Substituting back**:
Substitute \( b^2 \) into Equation 1:
\[ \frac{4}{a^2} + \frac{4}{\frac{3}{5}a^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{20}{3a^2} = 1 \]
Finding a common denominator:
\[ \frac{12 + 20}{3a^2} = 1 \implies \frac{32}{3a^2} = 1 \]
Thus:
\[ 3a^2 = 32 \implies a^2 = \frac{32}{3} \]

5. **Finding \( b^2 \)**:
Substitute \( a^2 \) back into \( b^2 \):
\[ b^2 = \frac{3}{5} \cdot \frac{32}{3} = \frac{32}{5} \]

6. **Final equation of the ellipse**:
Substitute \( a^2 \) and \( b^2 \) into the standard form:
\[ \frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1 \]
Multiplying through by 15 gives:
\[ 5x^2 + 3y^2 = 32 \]

### Final Answers:
1. For part (i): \( 3x^2 + 12y^2 = 16 \)
2. For part (ii): \( 5x^2 + 3y^2 = 32 \)