Find the eqation of the ellipse having foci (0,1),(0,-1) and minor axis of length 1.

Answer

To find the equation of the ellipse with foci at (0, 1) and (0, -1) and a minor axis of length 1, we can follow these steps:

### Step 1: Identify the coordinates of the foci and the minor axis
The foci of the ellipse are given as (0, 1) and (0, -1). This indicates that the ellipse is vertically oriented. The minor axis length is given as 1.

### Step 2: Determine the values of b and a
The distance from the center to each focus (c) is equal to 1 (since the foci are at (0, ±1)), so we have:
- \( c = 1 \)

The minor axis length is given as 1, which means:
- \( 2a = 1 \)
- Therefore, \( a = \frac{1}{2} \)

### Step 3: Use the relationship between a, b, and c
For an ellipse, the relationship between the semi-major axis (a), semi-minor axis (b), and the distance to the foci (c) is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
- \( c^2 = 1^2 = 1 \)
- \( a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

Now we can substitute these into the equation:
\[ 1 = \frac{1}{4} + b^2 \]

### Step 4: Solve for b
Rearranging the equation to solve for \( b^2 \):
\[ b^2 = 1 - \frac{1}{4} \]
\[ b^2 = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]
Thus, we find:
\[ b = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 5: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with a vertical major axis is given by:
\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]
Substituting the values of \( a \) and \( b \):
\[ \frac{x^2}{\frac{3}{4}} + \frac{y^2}{\frac{1}{4}} = 1 \]

### Step 6: Simplify the equation
To eliminate the fractions, we can multiply the entire equation by 4:
\[ 4 \cdot \frac{x^2}{\frac{3}{4}} + 4 \cdot \frac{y^2}{\frac{1}{4}} = 4 \]
This simplifies to:
\[ \frac{16x^2}{3} + 4y^2 = 4 \]

### Step 7: Final equation
Rearranging gives us:
\[ 16x^2 + 12y^2 = 12 \]
Dividing through by 12:
\[ \frac{4x^2}{3} + y^2 = 1 \]

Thus, the final equation of the ellipse is:
\[ \frac{4x^2}{3} + y^2 = 1 \]