Find the equation of the ellipse whose centre is at (0, 2) and major axis along the axis of y and whose minor axis is equal to the distance between the foci and whose latus rectum is 2.

Answer

To find the equation of the ellipse given the conditions, we can follow these steps:

### Step 1: Identify the standard form of the ellipse
Since the major axis is along the y-axis and the center is at (0, 2), the standard form of the ellipse can be written as:
\[ \frac{x^2}{b^2} + \frac{(y - 2)^2}{a^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Use the given information about the latus rectum
The length of the latus rectum \( L \) for an ellipse is given by:
\[ L = \frac{2a^2}{b} \]
We are told that the latus rectum is 2, so we can set up the equation:
\[ \frac{2a^2}{b} = 2 \]
From this, we can simplify to find:
\[ a^2 = b \]

### Step 3: Use the relationship between the distance between the foci and the minor axis
The distance between the foci \( 2c \) for an ellipse is given by:
\[ c = \sqrt{a^2 - b^2} \]
We know that the distance between the foci is equal to the minor axis, which is \( 2b \). Therefore, we have:
\[ 2c = 2b \implies c = b \]
Substituting for \( c \):
\[ b = \sqrt{a^2 - b^2} \]
Squaring both sides gives:
\[ b^2 = a^2 - b^2 \]
Rearranging this, we find:
\[ 2b^2 = a^2 \implies a^2 = 2b^2 \]

### Step 4: Substitute \( a^2 \) from Step 2 into Step 3
From Step 2, we have \( a^2 = b \). Substituting this into \( a^2 = 2b^2 \):
\[ b = 2b^2 \]
This can be rearranged to:
\[ 2b^2 - b = 0 \implies b(2b - 1) = 0 \]
This gives us \( b = 0 \) or \( b = \frac{1}{2} \). Since \( b \) cannot be zero, we have:
\[ b = \frac{1}{2} \]

### Step 5: Find \( a \) using \( a^2 = b \)
Now substituting \( b = \frac{1}{2} \) back into \( a^2 = b \):
\[ a^2 = \frac{1}{2} \]

### Step 6: Write the equation of the ellipse
Now we can substitute \( a^2 \) and \( b^2 \) into the standard form:
\[ \frac{x^2}{\left(\frac{1}{2}\right)} + \frac{(y - 2)^2}{\left(\frac{1}{2}\right)} = 1 \]
Multiplying through by 2 to eliminate the fractions gives:
\[ 2x^2 + 2(y - 2)^2 = 2 \]
Expanding this:
\[ 2x^2 + 2(y^2 - 4y + 4) = 2 \]
Simplifying further:
\[ 2x^2 + 2y^2 - 8y + 8 = 2 \]
Subtracting 2 from both sides:
\[ 2x^2 + 2y^2 - 8y + 6 = 0 \]

### Final Equation
Thus, the final equation of the ellipse is:
\[ 2x^2 + 2y^2 - 8y + 6 = 0 \]

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