The equation of the hyperbola with its transverse axis parallel to x-axis and its centre is ( -2,1) the length of transverse axis is 10 and eccentricity 6/5 is

Options

A. `x^(2)-2y^(2)+18x+27=0`
B. `((x+2)^(2))/(25)-((y-1)^(2))/(11)=1`
C. `((x-3)^(2))/(6)-((y-2)^(2))/(9)=1`
D. `((x-2)^(2))/(16)-((y-3)^(2))/(19)=1`

Answer

To find the equation of the hyperbola given the conditions, we can follow these steps:

### Step 1: Identify the center of the hyperbola
The center of the hyperbola is given as \((-2, 1)\). This means \(h = -2\) and \(k = 1\).

### Step 2: Determine the length of the transverse axis
The length of the transverse axis is given as 10. The length of the transverse axis is equal to \(2a\), where \(a\) is the distance from the center to the vertices along the transverse axis.

\[ 2a = 10 \implies a = \frac{10}{2} = 5 \]

### Step 3: Use the eccentricity to find \(c\)
The eccentricity \(e\) is given as \(\frac{6}{5}\). The relationship between eccentricity, \(c\), and \(a\) is given by:

\[ e = \frac{c}{a} \]

Substituting the known values:

\[ \frac{6}{5} = \frac{c}{5} \]

Multiplying both sides by 5:

\[ c = 6 \]

### Step 4: Find \(b\) using the relationship between \(a\), \(b\), and \(c\)
We know that:

\[ c^2 = a^2 + b^2 \]

Substituting the values of \(c\) and \(a\):

\[ 6^2 = 5^2 + b^2 \]

Calculating the squares:

\[ 36 = 25 + b^2 \]

Rearranging gives:

\[ b^2 = 36 - 25 = 11 \]

### Step 5: Write the equation of the hyperbola
The standard form of the equation of a hyperbola with a transverse axis parallel to the x-axis and center at \((h, k)\) is:

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

Substituting \(h = -2\), \(k = 1\), \(a^2 = 25\), and \(b^2 = 11\):

\[ \frac{(x + 2)^2}{25} - \frac{(y - 1)^2}{11} = 1 \]

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

\[ \frac{(x + 2)^2}{25} - \frac{(y - 1)^2}{11} = 1 \]

---