The equation of the hyperola whose centre is (5,2) vertex is (9,2) and the length of conjugate axis is 6 is

Options

A. `((x-5)^(2))/(16)+((y-2)^(2))/(9)=1`
B. `((x-5)^(2))/(16)-((y-2)^(2))/(9)=1`
C. `((x-5)^(2))/(16)-((y-2)^(2))/(9)=-1`
D. `((x+5)^(2))/(26)-((y+2)^(2))/(29)=1`

Answer

To find the equation of the hyperbola with the given parameters, we will follow these steps:

### Step 1: Identify the center and vertex
The center of the hyperbola is given as \( (5, 2) \) and the vertex is at \( (9, 2) \).

### Step 2: Determine the value of \( a \)
The distance from the center to the vertex is equal to \( a \).

\[ a = \text{distance from center to vertex} = |9 - 5| = 4 \]

### Step 3: Find the length of the conjugate axis
The length of the conjugate axis is given as 6, which means \( 2b = 6 \). Therefore, we can find \( b \):

\[ b = \frac{6}{2} = 3 \]

### Step 4: Write the standard form of the hyperbola
Since the vertex is horizontal (the y-coordinates are the same), the equation of the hyperbola will be in the form:

\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]

Where \( (h, k) \) is the center of the hyperbola.

### Step 5: Substitute the values into the equation
Substituting \( h = 5 \), \( k = 2 \), \( a = 4 \), and \( b = 3 \):

\[ \frac{(x - 5)^2}{4^2} - \frac{(y - 2)^2}{3^2} = 1 \]

This simplifies to:

\[ \frac{(x - 5)^2}{16} - \frac{(y - 2)^2}{9} = 1 \]

### Final Answer
Thus, the equation of the hyperbola is:

\[ \frac{(x - 5)^2}{16} - \frac{(y - 2)^2}{9} = 1 \]

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