Find the equation of the hyperbola with :
vertices `(0,pm7), e=(7)/(3)`.

Answer

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

### Step 1: Identify the vertices and parameters
The vertices of the hyperbola are given as \( (0, \pm 7) \). This indicates that the hyperbola opens vertically. The distance from the center to the vertices is denoted by \( b \). Therefore, we have:
\[ b = 7 \]

### Step 2: Use the eccentricity formula
The eccentricity \( e \) of the hyperbola is given as \( \frac{7}{3} \). The relationship between eccentricity, \( a \), and \( b \) in a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{a^2}{b^2}} \]
Substituting the known values:
\[ \frac{7}{3} = \sqrt{1 + \frac{a^2}{7^2}} \]
This simplifies to:
\[ \frac{7}{3} = \sqrt{1 + \frac{a^2}{49}} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{7}{3}\right)^2 = 1 + \frac{a^2}{49} \]
Calculating the left side:
\[ \frac{49}{9} = 1 + \frac{a^2}{49} \]

### Step 4: Rearranging the equation
Subtract 1 from both sides:
\[ \frac{49}{9} - 1 = \frac{a^2}{49} \]
Convert 1 into a fraction with a denominator of 9:
\[ \frac{49}{9} - \frac{9}{9} = \frac{a^2}{49} \]
This simplifies to:
\[ \frac{40}{9} = \frac{a^2}{49} \]

### Step 5: Cross-multiply to solve for \( a^2 \)
Cross-multiplying gives:
\[ 40 \cdot 49 = 9 \cdot a^2 \]
Calculating \( 40 \cdot 49 \):
\[ 1960 = 9a^2 \]
Now, divide both sides by 9:
\[ a^2 = \frac{1960}{9} \]

### Step 6: Write the equation of the hyperbola
Since the hyperbola opens vertically, its standard form is:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]
Substituting \( b^2 = 49 \) and \( a^2 = \frac{1960}{9} \):
\[ \frac{y^2}{49} - \frac{x^2}{\frac{1960}{9}} = 1 \]

### Step 7: Simplify the equation
To eliminate the fraction in the second term, multiply through by \( 9 \):
\[ 9 \cdot \frac{y^2}{49} - 9 \cdot \frac{x^2}{\frac{1960}{9}} = 9 \]
This simplifies to:
\[ \frac{9y^2}{49} - \frac{9x^2 \cdot 9}{1960} = 9 \]
Thus, the final equation of the hyperbola is:
\[ \frac{9y^2}{49} - \frac{81x^2}{1960} = 1 \]