Step-by-step Solution:

The given equation is: \[ x^2 + 2x - 4y^2 + 8y - 7 = 0 \] Step 1: Identify the Conic Section
Rewriting the equation: \[ x^2 + 2x - 4(y^2 - 2y) = 7 \] Since the coefficients of \(x^2\) and \(y^2\) have opposite signs, it represents a hyperbola.
Step 2: Complete the Square
Completing the square for \(x\): \[ x^2 + 2x = (x + 1)^2 - 1 \] Completing the square for \(y\): \[ y^2 - 2y = (y - 1)^2 - 1 \] Multiplying by \(-4\): \[ -4(y^2 - 2y) = -4((y - 1)^2 - 1) = -4(y - 1)^2 + 4 \] Substituting these back into the equation: \[ (x + 1)^2 - 1 - 4(y - 1)^2 + 4 = 7 \] \[ (x + 1)^2 - 4(y - 1)^2 + 3 = 7 \] \[ (x + 1)^2 - 4(y - 1)^2 = 4 \] \[ \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 \] Step 3: Find the Foci
For the standard hyperbola equation: \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \] We compare:
\( h = -1 \), \( k = 1 \)
\( a^2 = 4 \Rightarrow a = 2 \)
\( b^2 = 1 \Rightarrow b = 1 \)
The focal distance is given by: \[ c^2 = a^2 + b^2 = 4 + 1 = 5 \] \[ c = \sqrt{5} \] Since it is a horizontally oriented hyperbola, the foci are at: \[ (x + 1) = \pm c \] \[ (x, y) = (-1 \pm \sqrt{5}, 1) \] Final Answer: \[ {(-1 + \sqrt{5}, 1) \text{ and } (-1 - \sqrt{5}, 1)} \]