Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To find a asymptote its either b2/a2 or a2/b2 depending on the way the equation is written.

With the problem

$$\frac{(x+1)^2}{16} - \frac{(y-2)^2}{9} = 1$$

The solutions the sheet I have is giving me is $3/4x - 3/4$ and $3/4 x + 5/4$

I thought it was just supposed to be $\pm 3/4x$.

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

For the hyperbola $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$ The asymptotes are $y - k = \pm \dfrac{b}{a}(x - h)$.

You could leave your answer as $y - 2 = \pm \dfrac{3}{4}(x + 1)$, or write two separate equations.


Edit... If you do write separate equations, you'll have

$y - 2 = \dfrac{3}{4}(x + 1)$ and $y - 2 = - \dfrac{3}{4}(x + 1)$, which are, in "slope-intercept form":

$y = \dfrac{3}{4} x + \dfrac{11}{4}$ and $y = - \dfrac{3}{4}x + \dfrac{5}{4}$

share|improve this answer
    
thats what I thought, but the sheet I'm working off of has -3/4 x + 5/4 as a solution. I'm confused to as of where they got the 5/4 Every learning resource I've looked at just says b/a. –  CrewdNBasic Jul 7 '12 at 0:58
    
Not to be pedantic, but the resources certainly do not say "the answer is b/a". The resources probably say exactly what my second line says: "The asymptotes are $y - k = \pm \cdots$." Five fourths is the y-intercept of one of the asymptotes. –  The Chaz 2.0 Jul 7 '12 at 1:10
    
Thanks...so why is the solution listed as y = 3/4x - 3/4 and -3/4x +5/4 instead of y = 3/4+11/4 and y = -3/4x +5/4 –  CrewdNBasic Jul 7 '12 at 1:19
    
Either I have made an arithmetic mistake, or the solution author has (or both!). –  The Chaz 2.0 Jul 7 '12 at 1:30
    
Nope. Yours is right because add 2, multiply it by 4 because of the denominator and add to 3 and you should get 11/4. –  CrewdNBasic Jul 7 '12 at 1:40
add comment

For hyperbola $(x+1)^2/16 - (y-2)^2/9 = 1$, the equation for the asymptotes is $(x+1)^2/16 - (y-2)^2/9 = 0$. This can be factored into two linear equations, corresponding to two lines. The center of your hyperbola is $(-1,2)$, so of course the two asymptotes go through that point.

share|improve this answer
add comment

Hint: Your hyperbola is the standard hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, moved one unit to the left and $2$ units up. So write down the asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$, and move them in the same way.

Added: The asymptotes of $\frac{x^2}{16}-\frac{y^2}{9}=1$ are, as you know, $\frac{y}{3}=\pm \frac{x}{4}$. So the asymptotes in your case are $$\frac{y-2}{3}=\pm\frac{x+1}{4}.$$ The "plus" case simplifies to $y-2=\frac{3}{4}x+\frac{3}{4}$, then to $y=\frac{3}{4}x+\frac{11}{4}$.

The "minus" case simplifies to $y=-\frac{3}{4}x+\frac{5}{4}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.