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 eccentricity
The vertices of the hyperbola are given as \( (0, \pm 7) \) and the eccentricity \( e = \frac{7}{3} \).

### Step 2: Determine the value of \( b \)
From the vertices, we know that the hyperbola opens vertically (since the vertices are along the y-axis). The standard form of the equation of a hyperbola that opens vertically is:

\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

Here, the vertices are \( (0, \pm a) \), so we have \( a = 7 \).

### Step 3: Calculate \( a^2 \)
Now, we calculate \( a^2 \):

\[ a^2 = 7^2 = 49 \]

### Step 4: Use the eccentricity to find \( b^2 \)
The relationship between the eccentricity \( e \), \( a \), and \( b \) is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Squaring both sides gives:

\[ e^2 = 1 + \frac{b^2}{a^2} \]

Substituting the values we have:

\[ \left(\frac{7}{3}\right)^2 = 1 + \frac{b^2}{49} \]

Calculating \( \left(\frac{7}{3}\right)^2 \):

\[ \frac{49}{9} = 1 + \frac{b^2}{49} \]

### Step 5: Solve for \( b^2 \)
Rearranging the equation gives:

\[ \frac{49}{9} - 1 = \frac{b^2}{49} \]

Converting 1 to a fraction with a denominator of 9:

\[ \frac{49}{9} - \frac{9}{9} = \frac{b^2}{49} \]

This simplifies to:

\[ \frac{40}{9} = \frac{b^2}{49} \]

Cross-multiplying gives:

\[ 40 \cdot 49 = 9b^2 \]

Calculating \( 40 \cdot 49 \):

\[ 1960 = 9b^2 \]

Now, dividing both sides by 9:

\[ b^2 = \frac{1960}{9} \]

### Step 6: Write the equation of the hyperbola
Now we have \( a^2 = 49 \) and \( b^2 = \frac{1960}{9} \). Substituting these values into the standard form of the hyperbola:

\[ \frac{y^2}{49} - \frac{x^2}{\frac{1960}{9}} = 1 \]

To eliminate the fraction in the denominator of the second term, we can multiply the entire equation by 9:

\[ 9 \cdot \frac{y^2}{49} - 9 \cdot \frac{x^2}{\frac{1960}{9}} = 9 \]

This simplifies to:

\[ \frac{9y^2}{49} - \frac{9x^2}{\frac{1960}{9}} = 9 \]

Rearranging gives:

\[ 9y^2 - \frac{9 \cdot 49}{1960} x^2 = 441 \]

Multiplying through by 1960 gives:

\[ 1960 \cdot 9y^2 - 9 \cdot 49x^2 = 441 \cdot 1960 \]

This results in the final equation of the hyperbola:

\[ 9x^2 - 40y^2 = 1960 \]

### Final Equation
The equation of the hyperbola is:

\[ 9x^2 - 40y^2 = 1960 \]