The equation of the hyperbola with verticles `(0, pm7)` and eccentricity `=(9)/(7)` is

Options

A. `(x^(2))/(49) - (y^(2))/(32) =1`
B. `(y^(2))/(49) - (x^(2))/(32) =1`
C. `(x^(2))/(7) - (y^(2))/(4sqrt(2)) =1`
D. `(y^(2))/(7) - (x^(2))/(4sqrt(2)) =1`

Answer

To find the equation of the hyperbola with given vertices and eccentricity, we can follow these steps:

### Step 1: Identify the vertices and eccentricity
The vertices of the hyperbola are given as (0, ±7). This means that the hyperbola is oriented vertically. The eccentricity \( e \) is given as \( \frac{9}{7} \).

### Step 2: Determine the value of \( a \)
For a hyperbola, the distance from the center to the vertices is denoted by \( a \). Since the vertices are at (0, ±7), we have:
\[ a = 7 \]

### Step 3: Calculate \( e^2 \)
The eccentricity \( e \) is related to \( a \) and \( b \) (the distance from the center to the foci) by the formula:
\[ e = \frac{c}{a} \]
where \( c \) is the distance from the center to the foci. We can find \( c \) using the eccentricity:
\[ c = e \cdot a = \frac{9}{7} \cdot 7 = 9 \]

### Step 4: Use the relationship \( c^2 = a^2 + b^2 \)
We know that:
\[ c^2 = a^2 + b^2 \]
Substituting the values we have:
\[ 9^2 = 7^2 + b^2 \]
\[ 81 = 49 + b^2 \]
\[ b^2 = 81 - 49 = 32 \]

### Step 5: Write the standard form of the hyperbola
The standard form of the equation of a hyperbola centered at the origin with vertical transverse axis is given by:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
Substituting \( a^2 = 49 \) and \( b^2 = 32 \):
\[ \frac{y^2}{49} - \frac{x^2}{32} = 1 \]

### Final Equation
Thus, the equation of the hyperbola is:
\[ \frac{y^2}{49} - \frac{x^2}{32} = 1 \]