`x^(2) -y^(2) +5x +8y -4=0 ` represents

Options

A. rectangular hyperbola
B. ellipse
C. hyperbola
D. pair of lines

Answer

To determine the type of curve represented by the equation \( x^2 - y^2 + 5x + 8y - 4 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ x^2 - y^2 + 5x + 8y - 4 = 0 \]
Rearranging it gives:
\[ x^2 + 5x - y^2 + 8y - 4 = 0 \]

### Step 2: Group the x and y terms
Group the x terms and the y terms:
\[ (x^2 + 5x) - (y^2 - 8y) - 4 = 0 \]

### Step 3: Complete the square for x terms
To complete the square for \( x^2 + 5x \):
1. Take half of the coefficient of \( x \) (which is \( 5 \)), square it: \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
2. Add and subtract \( \frac{25}{4} \):
\[ x^2 + 5x = \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} \]

### Step 4: Complete the square for y terms
To complete the square for \( -y^2 + 8y \):
1. Take half of the coefficient of \( y \) (which is \( 8 \)), square it: \( \left(\frac{8}{2}\right)^2 = 16 \).
2. Add and subtract \( 16 \):
\[ -y^2 + 8y = -\left(y^2 - 8y + 16\right) + 16 = -(y - 4)^2 + 16 \]

### Step 5: Substitute back into the equation
Substituting back into the equation gives:
\[ \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} - (y - 4)^2 + 16 - 4 = 0 \]
This simplifies to:
\[ \left(x + \frac{5}{2}\right)^2 - (y - 4)^2 + \frac{25}{4} + 12 = 0 \]
\[ \left(x + \frac{5}{2}\right)^2 - (y - 4)^2 + \frac{49}{4} = 0 \]

### Step 6: Rearranging to standard form
Rearranging gives:
\[ \left(x + \frac{5}{2}\right)^2 - (y - 4)^2 = -\frac{49}{4} \]
This can be rewritten as:
\[ \frac{\left(x + \frac{5}{2}\right)^2}{\frac{49}{4}} - \frac{(y - 4)^2}{\frac{49}{4}} = 1 \]

### Conclusion
The equation is now in the standard form of a hyperbola:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
where \( h = -\frac{5}{2} \), \( k = 4 \), \( a^2 = \frac{49}{4} \), and \( b^2 = \frac{49}{4} \).

Thus, the given equation represents a hyperbola.