Find the equation of the ellipse with its centre at (3, 1), vertex at (3, -2), and eccentricity equal to `1/3`.

Answer

To find the equation of the ellipse with its center at (3, 1), vertex at (3, -2), and eccentricity equal to \( \frac{1}{3} \), we can follow these steps:

### Step 1: Identify the center and vertex
The center of the ellipse is given as \( (h, k) = (3, 1) \). The vertex is at \( (3, -2) \).

### Step 2: Determine the distance from the center to the vertex (a)
The distance \( a \) from the center to the vertex can be calculated as follows:
\[ a = |k - y_{\text{vertex}}| = |1 - (-2)| = |1 + 2| = 3 \]

### Step 3: Find the value of \( a^2 \)
Now, we calculate \( a^2 \):
\[ a^2 = 3^2 = 9 \]

### Step 4: Use the eccentricity to find \( b^2 \)
The eccentricity \( e \) is given as \( \frac{1}{3} \). The relationship between \( a \), \( b \), and \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Squaring both sides gives:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting the values we have:
\[ \left(\frac{1}{3}\right)^2 = 1 - \frac{b^2}{9} \]
This simplifies to:
\[ \frac{1}{9} = 1 - \frac{b^2}{9} \]
Rearranging gives:
\[ \frac{b^2}{9} = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus, multiplying both sides by 9:
\[ b^2 = 8 \]

### Step 5: Write the equation of the ellipse
Since the vertex is vertical (the y-coordinate changes), the standard form of the ellipse is:
\[ \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \]
Substituting \( h = 3 \), \( k = 1 \), \( a^2 = 9 \), and \( b^2 = 8 \):
\[ \frac{(x - 3)^2}{8} + \frac{(y - 1)^2}{9} = 1 \]

### Step 6: Rearranging the equation
To express the equation in a more standard form, we can leave it as:
\[ \frac{(x - 3)^2}{8} + \frac{(y - 1)^2}{9} = 1 \]

### Final Equation
The equation of the ellipse is:
\[ \frac{(x - 3)^2}{8} + \frac{(y - 1)^2}{9} = 1 \]

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