Question

Find foci of the equation \( x^2 + 2x - 4y^2 + 8y - 7 = 0 \)

• A. \( (\sqrt{5} \pm 1, 1) \)
• B. \( (-1 \pm \sqrt{5}, 1) \)
• C. \( (-1 \pm \sqrt{5}, \pm 1) \)
• D. \( (1, -1 \pm \sqrt{5}) \)

Hint:

For hyperbolas, use the standard form \( \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \) to find the foci, where \( c^2 = a^2 + b^2 \).

Answer 1

The Correct Option is B

Step 1: The given equation is:
\[ x^2 + 2x - 4y^2 + 8y - 7 = 0 \] Step 2: Rearrange the terms:
\[ (x^2 + 2x) - (4y^2 - 8y) = 7 \] Step 3: Complete the square for both \( x \) and \( y \): - For \( x^2 + 2x \), add and subtract \( 1 \) to complete the square:
\[ (x^2 + 2x + 1) = (x + 1)^2 \] - For \( -4y^2 + 8y \), factor out the \( -4 \):
\[ -4(y^2 - 2y) = -4(y^2 - 2y + 1 - 1) = -4((y - 1)^2 - 1) \] Thus, the equation becomes:
\[ (x + 1)^2 - 4((y - 1)^2 - 1) = 7 \] Simplifying further:
\[ (x + 1)^2 - 4(y - 1)^2 + 4 = 7 \] \[ (x + 1)^2 - 4(y - 1)^2 = 3 \] Step 4: Divide by 3 to get the standard form:
\[ \frac{(x + 1)^2}{3} - \frac{(y - 1)^2}{\frac{3}{4}} = 1 \] Step 5: This is the standard form of a hyperbola with center \( (-1, 1) \), \( a^2 = 3 \), and \( b^2 = \frac{3}{4} \). The foci are given by \( \pm \sqrt{a^2 + b^2} \) along the \( x \)-axis (since the hyperbola opens left-right).
\[ c^2 = a^2 + b^2 = 3 + \frac{3}{4} = \frac{15}{4} \] \[ c = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \] Thus, the foci are at \( (-1 \pm \sqrt{5}, 1) \). Therefore, the correct answer is (b) \( (-1 \pm \sqrt{5}, 1) \).