Find the equation of the hyperbola with vertices at `(0,+-6)` and `e=5/3`

Answer

To find the equation of the hyperbola with vertices at (0, ±6) and eccentricity \( e = \frac{5}{3} \), we can follow these steps:

### Step 1: Identify the values of \( b \)
The vertices of the hyperbola are given as (0, ±6). In the standard form of the hyperbola centered at the origin, the vertices are located at (0, ±b). Therefore, we have:
\[ b = 6 \]

### Step 2: Calculate \( b^2 \)
We need to find \( b^2 \):
\[ b^2 = 6^2 = 36 \]

### Step 3: Use the relationship between \( e \), \( a \), and \( b \)
The eccentricity \( e \) of a hyperbola is related to \( a \) and \( b \) by the formula:
\[ e = \frac{c}{a} \]
where \( c = \sqrt{a^2 + b^2} \). We can also express this as:
\[ e^2 = \frac{c^2}{a^2} = \frac{a^2 + b^2}{a^2} \]
From this, we can derive:
\[ e^2 = 1 + \frac{b^2}{a^2} \]

### Step 4: Substitute the known values into the equation
Given \( e = \frac{5}{3} \), we can calculate \( e^2 \):
\[ e^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]
Now, substituting \( b^2 = 36 \) into the equation:
\[ \frac{25}{9} = 1 + \frac{36}{a^2} \]

### Step 5: Rearrange the equation to solve for \( a^2 \)
Rearranging gives:
\[ \frac{25}{9} - 1 = \frac{36}{a^2} \]
Converting 1 to a fraction with a denominator of 9:
\[ \frac{25}{9} - \frac{9}{9} = \frac{36}{a^2} \]
This simplifies to:
\[ \frac{16}{9} = \frac{36}{a^2} \]

### Step 6: Cross-multiply to solve for \( a^2 \)
Cross-multiplying gives:
\[ 16a^2 = 36 \cdot 9 \]
Calculating the right side:
\[ 16a^2 = 324 \]
Now, divide both sides by 16:
\[ a^2 = \frac{324}{16} = 20.25 \]

### Step 7: Write the equation of the hyperbola
The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \]
Substituting \( a^2 = 20.25 \) and \( b^2 = 36 \):
\[ \frac{x^2}{20.25} - \frac{y^2}{36} = -1 \]

### Final Answer
Thus, the equation of the hyperbola is:
\[ \frac{x^2}{20.25} - \frac{y^2}{36} = -1 \]