Question

The equation of the pair of asymptotes of the hyperbola $x^2 - 2y^2 - 8x + 8y + 4 = 0$ is

• A. $x^2 - 2y^2 - 8x + 8y - 8 = 0$
• B. $2x^2 - 4y^2 - 16x + 16y - 7 = 0$
• C. $x^2 - 2y^2 - 8x + 8y + 8 = 0$
• D. $2x^2 - 4y^2 - 16x + 16y + 9 = 0$

Hint:

Hyperbola and Asymptotes:

• General conic: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$
• Asymptotes differ from conic only in the constant term.
• Use partial derivatives $\partial H/\partial x$, $\partial H/\partial y$ to find center.

Answer 1

The Correct Option is C

Given hyperbola $H: x^2 - 2y^2 - 8x + 8y + 4 = 0$ Let asymptote equation be $A: x^2 - 2y^2 - 8x + 8y + k = 0$ Find center: \[ \frac{\partial H}{\partial x} = 2x - 8 = 0 \Rightarrow x = 4, \quad \frac{\partial H}{\partial y} = -4y + 8 = 0 \Rightarrow y = 2 \] Substitute $(4,2)$ into $A$: \[ 16 - 8 - 32 + 16 + k = 0 \Rightarrow k = 8 \] So asymptotes: $x^2 - 2y^2 - 8x + 8y + 8 = 0$

Answer 2

Step 1: Write the given hyperbola equation
The given equation is:
\[ x^2 - 2y^2 - 8x + 8y + 4 = 0 \]

Step 2: Find the center by completing the square
Group \(x\) and \(y\) terms:
\[ (x^2 - 8x) - 2(y^2 - 4y) + 4 = 0 \]
Complete the square for \(x\):
\[ x^2 - 8x = (x - 4)^2 - 16 \]
Complete the square for \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Substitute back:
\[ (x - 4)^2 - 16 - 2\big[(y - 2)^2 - 4\big] + 4 = 0 \]
Simplify:
\[ (x - 4)^2 - 16 - 2(y - 2)^2 + 8 + 4 = 0 \]\[ (x - 4)^2 - 2(y - 2)^2 - 4 = 0 \]

Step 3: Rearrange to standard hyperbola form
\[ (x - 4)^2 - 2(y - 2)^2 = 4 \]
Divide both sides by 4:
\[ \frac{(x - 4)^2}{4} - \frac{(y - 2)^2}{2} = 1 \]

Step 4: Write the equations of asymptotes
For hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), asymptotes are:
\[ y - k = \pm \frac{b}{a}(x - h) \]
Here, \(a^2 = 4\), \(b^2 = 2\), so
\[ y - 2 = \pm \frac{\sqrt{2}}{2} (x - 4) \]
Multiply both sides by 2:
\[ 2(y - 2) = \pm \sqrt{2}(x - 4) \]
Square both sides to eliminate the ±:
\[ 4(y - 2)^2 = 2(x - 4)^2 \]\[ 2(x - 4)^2 - 4(y - 2)^2 = 0 \]

Step 5: Expand and simplify asymptotes equation
Expand:
\[ 2(x^2 - 8x + 16) - 4(y^2 - 4y + 4) = 0 \]\[ 2x^2 - 16x + 32 - 4y^2 + 16y - 16 = 0 \]\[ 2x^2 - 16x - 4y^2 + 16y + 16 = 0 \]
Divide entire equation by 2:
\[ x^2 - 8x - 2y^2 + 8y + 8 = 0 \]

Conclusion:
The equation of the pair of asymptotes is:
\[ x^2 - 2y^2 - 8x + 8y + 8 = 0 \]