Question

For the hyperbola $\frac{y^2}{16}-\frac{x^2}{81}=1$, the asymptotes are _____ and _______

Short answer

For the hyperbola $\frac{y^2}{16}-\frac{x^2}{81}=1$, the asymptotes are $y=\frac{2}{3}x$and $y=\frac{-2}{3}x$.

Step-by-step solution

Step 1. Given Information

The equation of the hyperbola is $\frac{y^2}{16}-\frac{x^2}{81}=1$.

Step 2. Aim of the problem

Here, we need to find the asymptotes of the hyperbola$\frac{y^2}{16}-\frac{x^2}{81}=1$.

Step 3. Concept

The asymptotes of a hyperbola whose equation is of the form $\frac{{(y-k)}^2}{a^2}-\frac{{(x-h)}^2}{b^2}=1$is$y-k=\pm \frac{a}{b}(x-h)$.

Step 4. Finding the asymptotes

The given equation $\frac{y^2}{16}-\frac{x^2}{81}=1$is of the form $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$, where $(h,k)=(0,0)$and $a^2=16$and $b^2=81$.

Since, $a^2=16,$we have, $a=\sqrt{16}=4$and $b=\sqrt{81}=9$.

Therefore, the asymptotes are:

role="math" $$\begin{gathered} y-k=\pm \frac{a}{b}(x-h) \\ y-0=\pm \frac{4}{9}(x-0) \\ y=\pm \frac{2}{3}x \\ y=\frac{2}{3}xory=\frac{-2}{3}x \end{gathered}$$

Therefore, the asymptotes of the hyperbola are$y=\frac{2}{3}x$and$y=\frac{-2}{3}x$