Find the equation to the ellipse with axes as the axes of coordinates.
which passes through the points (3, -1) and (2, -2).

Answer

To find the equation of the ellipse with axes along the coordinate axes that passes through the points (3, -1) and (2, -2), we can follow these steps:

### Step 1: Write the standard form of the ellipse equation
The standard form of the ellipse with axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the first point (3, -1) into the equation
Substituting \(x = 3\) and \(y = -1\) into the ellipse equation:
\[ \frac{3^2}{a^2} + \frac{(-1)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (2, -2) into the equation
Now, substitute \(x = 2\) and \(y = -2\) into the ellipse equation:
\[ \frac{2^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the two equations simultaneously
From Equation 1:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \]
Multiply through by \(b^2\):
\[ 9b^2 + a^2 = b^2 \quad \Rightarrow \quad 9b^2 + a^2 = b^2 \quad \Rightarrow \quad a^2 = b^2 - 9b^2 \quad \Rightarrow \quad a^2 = -8b^2 \]
This is incorrect; let's go back to the equations.

From Equation 2:
\[ \frac{4}{a^2} + \frac{4}{b^2} = 1 \]
Multiply through by \(a^2b^2\):
\[ 4b^2 + 4a^2 = a^2b^2 \quad \Rightarrow \quad a^2b^2 - 4a^2 - 4b^2 = 0 \]

### Step 5: Solve for \(a^2\) and \(b^2\)
From Equation 1:
\[ \frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \Rightarrow \quad \frac{1}{b^2} = 1 - \frac{9}{a^2} \quad \Rightarrow \quad b^2 = \frac{a^2}{a^2 - 9} \]

Substituting \(b^2\) into Equation 2:
\[ \frac{4}{a^2} + \frac{4(a^2 - 9)}{a^2} = 1 \]
This gives:
\[ 4 + 4(a^2 - 9) = a^2 \]
Rearranging:
\[ 4a^2 - 36 = a^2 \quad \Rightarrow \quad 3a^2 = 36 \quad \Rightarrow \quad a^2 = 12 \]

### Step 6: Find \(b^2\)
Substituting \(a^2 = 12\) back into Equation 1:
\[ \frac{9}{12} + \frac{1}{b^2} = 1 \quad \Rightarrow \quad \frac{3}{4} + \frac{1}{b^2} = 1 \quad \Rightarrow \quad \frac{1}{b^2} = \frac{1}{4} \quad \Rightarrow \quad b^2 = 4 \]

### Step 7: Write the equation of the ellipse
Now we have \(a^2 = 12\) and \(b^2 = 4\). Thus, the equation of the ellipse is:
\[ \frac{x^2}{12} + \frac{y^2}{4} = 1 \]

### Step 8: Rearranging the equation
Multiplying through by 12 to eliminate the denominators:
\[ x^2 + 3y^2 = 12 \]

### Final Answer
Thus, the equation of the ellipse is:
\[ \boxed{x^2 + 3y^2 = 12} \]