How do you find the equations of the asymptotes of this hyperbola?

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

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- May 30th 2009, 08:36 AMjumpman23Asymptotes 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, 10:55 AMearboth
1. By completing the square you get:

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

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 $\displaystyle m_1 = 1~\vee~m_2=-1$

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

$\displaystyle a_1: y= x+5$

$\displaystyle a_2: y = -x +11$ - May 31st 2009, 07:25 AMHallsofIvy
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 $\displaystyle -\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 1$ 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 $\displaystyle -\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 0$ from which we get $\displaystyle \frac{(x- 3)^2}{64}= \frac{(y- 8)^2}{64}$ and then $\displaystyle x- 3= \pm (y- 8)$. 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.