The equation of the hyperbola with is transverse axis parallel to x-axis and its centre is ` (3,-2) ` the length of axes, are 8,6 is

Answer

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

### Step 1: Identify the center and the lengths of the axes
The center of the hyperbola is given as \( (3, -2) \). The lengths of the transverse and conjugate axes are given as 8 and 6, respectively.

### Step 2: Determine the values of \( a \) and \( b \)
The length of the transverse axis is \( 2a \), so:
\[ 2a = 8 \implies a = \frac{8}{2} = 4 \]

The length of the conjugate axis is \( 2b \), so:
\[ 2b = 6 \implies b = \frac{6}{2} = 3 \]

### Step 3: Write the standard form of the hyperbola
Since the transverse axis is parallel to the x-axis, the standard form of the equation of the hyperbola is:
\[ \frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1 \]
where \( (h, k) \) is the center of the hyperbola.

### Step 4: Substitute the values into the equation
Substituting \( h = 3 \), \( k = -2 \), \( a = 4 \), and \( b = 3 \):
\[ \frac{(y + 2)^2}{3^2} - \frac{(x - 3)^2}{4^2} = 1 \]
This simplifies to:
\[ \frac{(y + 2)^2}{9} - \frac{(x - 3)^2}{16} = 1 \]

### Step 5: Final equation of the hyperbola
Thus, the equation of the hyperbola is:
\[ \frac{(y + 2)^2}{9} - \frac{(x - 3)^2}{16} = 1 \]

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