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: $x^2 - 2y^2 - 8x + 8y + 8 = 0$