Answer :
To determine the equation of the hyperbola, follow these steps carefully:
### Step 1: Identify Hyperbola Characteristics
Given:
- Foci: \((7, 5)\) and \((7, -5)\)
- Directrix: \(y = \frac{16}{5}\)
From the coordinates of the foci:
- The foci are vertically aligned at \( x = 7 \).
- This indicates that the hyperbola opens vertically, meaning its equation has the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).
### Step 2: Determine the Center
The center \((h, k)\) of the hyperbola lies at the midpoint between the foci, which is calculated as:
[tex]\[ h = 7, \quad k = 0 \][/tex]
Thus, the center is at \((7, 0)\).
### Step 3: Calculate Distance \(c\)
The distance to each focus from the center is represented by \(c\). Given the coordinates of the foci, the distance \(c\) is:
[tex]\[ c = 5 \][/tex]
### Step 4: Determine \(a\) Using the Directrix
The directrix \( y = \frac{16}{5} \) aids in finding the value of \(a\).
For vertical hyperbolas, \(|k - \text{directrix}_y| = \frac{a^2}{c}\):
[tex]\[ \left| 0 - \frac{16}{5} \right| = \frac{a^2}{5} \][/tex]
[tex]\[ \frac{16}{5} = \frac{a^2}{5} \][/tex]
Solving for \(a\):
[tex]\[ a^2 = \frac{16}{5} \cdot 5 = 16 \][/tex]
Thus:
[tex]\[ a = 4 \][/tex]
### Step 5: Determine \(b\) Using the Relationship
For hyperbolas, the relationship \(c^2 = a^2 + b^2\) holds true:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Given:
[tex]\[ 25 = 16 + b^2 \][/tex]
Solving for \(b^2\):
[tex]\[ b^2 = 25 - 16 = 9 \][/tex]
Therefore:
[tex]\[ b = 3 \][/tex]
### Step 6: Form the Equation
Given \(a^2 = 16\) and \(b^2 = 9\), the standard form for the hyperbola with these values is:
[tex]\[ \frac{(y-0)^2}{16} - \frac{(x-7)^2}{9} = 1 \][/tex]
Simplified to:
[tex]\[ \frac{y^2}{16} - \frac{(x-7)^2}{9} = 1 \][/tex]
Therefore, the correct equation of the hyperbola is:
[tex]\[ \boxed{\frac{y^2}{16} - \frac{(x-7)^2}{9} = 1} \][/tex]
### Step 1: Identify Hyperbola Characteristics
Given:
- Foci: \((7, 5)\) and \((7, -5)\)
- Directrix: \(y = \frac{16}{5}\)
From the coordinates of the foci:
- The foci are vertically aligned at \( x = 7 \).
- This indicates that the hyperbola opens vertically, meaning its equation has the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).
### Step 2: Determine the Center
The center \((h, k)\) of the hyperbola lies at the midpoint between the foci, which is calculated as:
[tex]\[ h = 7, \quad k = 0 \][/tex]
Thus, the center is at \((7, 0)\).
### Step 3: Calculate Distance \(c\)
The distance to each focus from the center is represented by \(c\). Given the coordinates of the foci, the distance \(c\) is:
[tex]\[ c = 5 \][/tex]
### Step 4: Determine \(a\) Using the Directrix
The directrix \( y = \frac{16}{5} \) aids in finding the value of \(a\).
For vertical hyperbolas, \(|k - \text{directrix}_y| = \frac{a^2}{c}\):
[tex]\[ \left| 0 - \frac{16}{5} \right| = \frac{a^2}{5} \][/tex]
[tex]\[ \frac{16}{5} = \frac{a^2}{5} \][/tex]
Solving for \(a\):
[tex]\[ a^2 = \frac{16}{5} \cdot 5 = 16 \][/tex]
Thus:
[tex]\[ a = 4 \][/tex]
### Step 5: Determine \(b\) Using the Relationship
For hyperbolas, the relationship \(c^2 = a^2 + b^2\) holds true:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Given:
[tex]\[ 25 = 16 + b^2 \][/tex]
Solving for \(b^2\):
[tex]\[ b^2 = 25 - 16 = 9 \][/tex]
Therefore:
[tex]\[ b = 3 \][/tex]
### Step 6: Form the Equation
Given \(a^2 = 16\) and \(b^2 = 9\), the standard form for the hyperbola with these values is:
[tex]\[ \frac{(y-0)^2}{16} - \frac{(x-7)^2}{9} = 1 \][/tex]
Simplified to:
[tex]\[ \frac{y^2}{16} - \frac{(x-7)^2}{9} = 1 \][/tex]
Therefore, the correct equation of the hyperbola is:
[tex]\[ \boxed{\frac{y^2}{16} - \frac{(x-7)^2}{9} = 1} \][/tex]
