# Asymptotes of a Hyperbola

• May 30th 2009, 09:36 AM
jumpman23
Asymptotes of a Hyperbola
How do you find the equations of the asymptotes of this hyperbola?

-x^2 + y^2 + 6x - 16y - 9 = 0
• May 30th 2009, 11:55 AM
earboth
Quote:

Originally Posted by jumpman23
How do you find the equations of the asymptotes of this hyperbola?

-x^2 + y^2 + 6x - 16y - 9 = 0

1. By completing the square you get:

$-(x^2-6x+9) + (y^2-16y+64)=9-9+64~\implies~-\dfrac{(x-3)^2}{64}+\dfrac{(y-8)^2}{64}=1$

2. The center of the hyperbola is C(3, 8).
The semi-axes are a = 8 and b = 8

3. Therefore the asymptotes have the slopes $m_1 = 1~\vee~m_2=-1$

4. Use the point-slope-formual to get the equations of the asymptotes:

$a_1: y= x+5$

$a_2: y = -x +11$
• May 31st 2009, 08:25 AM
HallsofIvy
Here's another way of arguing. The "asymptotes" of a hyperbola are the lines the hyperbola approaches for very very large values of x and y.

The equation of the hyperbola is $-\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 1$ as Earboth said. For very very large x and y those squares will each be far larger! Large enough that that "1" on the right side is negligible compared to them: For very very large x and y, that equation is very very close to $-\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 0$ from which we get $\frac{(x- 3)^2}{64}= \frac{(y- 8)^2}{64}$ and then $x- 3= \pm (y- 8)$. Taking x- 3= y- 8, y= x+ 5. Taking x- 3= -(y- 8), x- 3= -y+ 8, y= -x+ 11, the answer that Earboth gave.