Finding Asymptotes of Hyperbolas

Question

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 $\pm 3/4x$.

Answer 1

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.

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

Answer 2

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 $y=-\frac{3}{4}x+\frac{5}{4}$.

Answer 3

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.