If the centre, vertex and focus of a hyperbola be (0,0), (4,0) and (6,0) respectively, then the equation of the hyperbola is

Options

A. `4x^(2) -5y^(2) = 8`
B. `4x^(2)-5y^(2) = 80`
C. `5x^(2)-4y^(2) =80`
D. `5x^(2)-4y^(2) =8`

Explanation

To find the equation of the hyperbola given the center, vertex, and focus, we can follow these steps:

### Step 1: Identify the center, vertex, and focus
Given:
- Center (h, k) = (0, 0)
- Vertex (a, 0) = (4, 0)
- Focus (ae, 0) = (6, 0)

### Step 2: Determine the value of 'a'
The distance from the center to the vertex is 'a'. Since the vertex is at (4, 0):
- a = 4

### Step 3: Determine the value of 'e'
The distance from the center to the focus is 'ae'. Since the focus is at (6, 0):
- ae = 6

Now, substituting the value of 'a':
- 4e = 6
- e = 6/4 = 3/2

### Step 4: Use the relationship between e, a, and b
For hyperbolas, the relationship between the eccentricity (e), semi-major axis (a), and semi-minor axis (b) is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting the known values:
\[ \frac{3}{2} = \sqrt{1 + \frac{b^2}{4^2}} \]
\[ \frac{3}{2} = \sqrt{1 + \frac{b^2}{16}} \]

### Step 5: Square both sides to eliminate the square root
\[ \left(\frac{3}{2}\right)^2 = 1 + \frac{b^2}{16} \]
\[ \frac{9}{4} = 1 + \frac{b^2}{16} \]

### Step 6: Rearranging the equation
Subtract 1 from both sides:
\[ \frac{9}{4} - 1 = \frac{b^2}{16} \]
\[ \frac{9}{4} - \frac{4}{4} = \frac{b^2}{16} \]
\[ \frac{5}{4} = \frac{b^2}{16} \]

### Step 7: Solve for b^2
Multiply both sides by 16:
\[ b^2 = 16 \cdot \frac{5}{4} \]
\[ b^2 = 4 \cdot 5 = 20 \]

### Step 8: Write the equation of the hyperbola
The standard form of the hyperbola centered at (0, 0) with a horizontal transverse axis is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

Substituting the values of \( a^2 \) and \( b^2 \):
- \( a^2 = 4^2 = 16 \)
- \( b^2 = 20 \)

Thus, the equation becomes:
\[ \frac{x^2}{16} - \frac{y^2}{20} = 1 \]

### Step 9: Multiply through by 80 to eliminate the denominators
\[ 5x^2 - 4y^2 = 80 \]

### Final Answer
The equation of the hyperbola is:
\[ 5x^2 - 4y^2 = 80 \]

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