Find the equation of the ellipse whose foci are `(0pm 3)` and `e=3/4`

Answer

To find the equation of the ellipse whose foci are at (0, ±3) and eccentricity \( e = \frac{3}{4} \), we can follow these steps:

### Step 1: Identify the values of the foci and eccentricity
The foci of the ellipse are given as \( (0, \pm 3) \). This means that the distance from the center to each focus, denoted as \( c \), is 3. The eccentricity \( e \) is given as \( \frac{3}{4} \).

### Step 2: Relate \( c \), \( a \), and \( b \)
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 the equation:
\[ c^2 = a^2 - b^2 \]
We also know that:
\[ e = \frac{c}{a} \]
Substituting the known values:
\[ \frac{3}{4} = \frac{c}{a} \]
This implies:
\[ c = \frac{3}{4}a \]

### Step 3: Substitute the value of \( c \)
Since we know \( c = 3 \), we can substitute this into the equation:
\[ 3 = \frac{3}{4}a \]
To find \( a \), multiply both sides by \( \frac{4}{3} \):
\[ a = 4 \]

### Step 4: Calculate \( b \)
Now we can use the relationship \( c^2 = a^2 - b^2 \):
\[ 3^2 = 4^2 - b^2 \]
This simplifies to:
\[ 9 = 16 - b^2 \]
Rearranging gives:
\[ b^2 = 16 - 9 = 7 \]

### Step 5: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with vertical major axis is:
\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]
Substituting the values we found:
\[ \frac{x^2}{7} + \frac{y^2}{16} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{7} + \frac{y^2}{16} = 1 \]

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