For a hyperbola with its foci on the $\displaystyle x$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the $\displaystyle x$ terms together and $\displaystyle y$ terms together.

$\displaystyle 9x^2-4y^2-162x-8y+689=0$

$\displaystyle 9x^2-162x-4y^2-8y+689=0$

Factor out $\displaystyle 9$ from the $\displaystyle x$ terms and $\displaystyle -4$ from the $\displaystyle y$ terms.

$\displaystyle 9(x^2-18x)-4(y^2+2y)+689=0$

Complete the squares. Remember to add the amount amount to both sides of the equation!

$\displaystyle 9(x^2-18x+81)-4(y^2+2y+1)+689=725$

Subtract $\displaystyle 689$ from both sides of the equation:

$\displaystyle 9(x^2-18x+81)-4(y^2+2y+1)=36$

Divide both sides by $\displaystyle 36$.

$\displaystyle \frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{9}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\displaystyle \frac{(x-9)^2}{4}-\frac{(y+1)^2}{9}=1$

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=2$ and $\displaystyle b=3$.

Thus, the slopes for its asymptotes are $\displaystyle m=\pm\frac{3}{2}$.

Plug in the center of the hyperbola into the point-slope form of a line to find the equations of the asymptotes. The center of the hyperbola is $\displaystyle (9, -1)$.

$\displaystyle y+1=\pm\frac{3}{2}(x-9)$

Now, simplify each equation. For the first equation,

$\displaystyle y+1=\frac{3}{2}(x-9)$

$\displaystyle y+1=\frac{3}{2}x-\frac{27}{2}$

$\displaystyle y=\frac{3}{2}x-\frac{29}{2}$

For the second equation,

$\displaystyle y+1=-\frac{3}{2}(x-9)$

$\displaystyle y+1=-\frac{3}{2}x+\frac{27}{2}$

$\displaystyle y=-\frac{3}{2}x-\frac{25}{2}$

For a hyperbola with its foci on the $\displaystyle x$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=12$ and $\displaystyle b=13$.

Thus, the slopes for its asymptotes are $\displaystyle m=\pm\frac{13}{12}$

Now, plug in the coordinates for the center of the hyperbola $\displaystyle (12, \sqrt3)$ into the point-slope form of the line to find the equations of the asymptotes.

$\displaystyle y-\sqrt3=\pm\frac{13}{12}(x-12)$

$\displaystyle y=\pm\frac{13}{12}(x-12)+\sqrt3$

For a hyperbola with its foci on the $\displaystyle x$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle 10$ and $\displaystyle b=6$.

Thus, the slopes for its asymptotes are $\displaystyle m=\pm\frac{6}{10}=\pm\frac{3}{5}$.

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at $\displaystyle (2, 6)$.

The first equation of the asymptote would be the following:

$\displaystyle y-6=\frac{3}{5}(x-2)$

$\displaystyle y-6=\frac{3}{5}x-\frac{6}{5}$

$\displaystyle y=\frac{3}{5}x+\frac{24}{5}$

The second equation of the asymptote would be the following:

$\displaystyle y-6=-\frac{3}{5}(x-2)$

$\displaystyle y-6=-\frac{3}{5}x+\frac{6}{5}$

$\displaystyle y=-\frac{3}{5}+\frac{36}{5}$

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=5$ and $\displaystyle b=3$.

Since the center is $\displaystyle (0, 0)$, the equations for its asymptotes are $\displaystyle y=\pm\frac{3}{5}x$.

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=15$ and $\displaystyle b=13$.

Thus, the slopes for its asymptotes are $\displaystyle \pm\frac{13}{15}$.

Now, plug in the center of the hyperbola, $\displaystyle (2, 9)$ into the point-slope form of a line to find the equations of the asymptotes.

For the first asymptote, 

$\displaystyle y-9=\frac{13}{15}(x-2)$

$\displaystyle y-9=\frac{13}{15}x-\frac{26}{15}$

$\displaystyle y=\frac{13}{15}x+\frac{109}{15}$

For the second asymptote,

$\displaystyle y-9=-\frac{13}{15}(x-2)$

$\displaystyle y-9=-\frac{13}{15}x+\frac{26}{15}$

$\displaystyle y=-\frac{13}{15}x+\frac{161}{15}$

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=6$ and $\displaystyle b=\sqrt8=2\sqrt2$.

Thus, the slopes for its asymptotes are $\displaystyle \pm\frac{2\sqrt2}{6}=\pm\frac{\sqrt2}{3}$.

Now, plug in the center of the hyperbola into the point-slope form of the equation fo the line to get the equations of the asymptotes.

The center of the hyperbola is $\displaystyle (-2, 1)$. The equations of the asymptotes are then:

$\displaystyle y-1=\pm\frac{\sqrt2}{3}(x+2)$

$\displaystyle y=\pm\frac{\sqrt2}{3}(x+2)+1$

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=20$ and $\displaystyle b=13$.

Thus, the slopes for its asymptotes are $\displaystyle \pm\frac{13}{20}$.

Now, plug in the center of the hyperbola into the point-slope form of an equation of a line to find the equations of the asymptotes. The center of the hyperbola is $\displaystyle (-\pi, 4)$.

The equations of the asymptotes can be then given by the following:

$\displaystyle y-4=\pm\frac{13}{20}(x+\pi)$

$\displaystyle y=\pm\frac{13}{20}(x+\pi)+4$

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the $\displaystyle x$ terms together and $\displaystyle y$ terms together.

$\displaystyle 36y^2-16x^2-144y-32x-448=0$

$\displaystyle 36y^2-144y-16x^2-32x-448=0$

Factor out $\displaystyle -16$ from the $\displaystyle x$ terms and $\displaystyle 36$ from the $\displaystyle y$ terms.

$\displaystyle 36(y^2-4y)-16(x^2+2x)-448=0$

Complete the squares. Remember to add the amount amount to both sides of the equation!

$\displaystyle 36(y^2-4y+4)-16(x^2+2x+1)-448=98$

Add $\displaystyle 448$ to both sides of the equation:

$\displaystyle 36(y^2-4y+4)-16(x^2+2x+1)=576$

Divide both sides by $\displaystyle 576$.

$\displaystyle \frac{y^2-4y+4}{16}-\frac{x^2+2x+1}{36}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\displaystyle \frac{(y-2)^2}{16}-\frac{(x+1)^2}{36}=1$

The slopes of this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=6$ and $\displaystyle b=4$.

Thus, the slopes for its asymptotes are $\displaystyle \pm\frac{4}{6}=\pm\frac{2}{3}$.

Now, plug in the center of the hyperbola, $\displaystyle (-1, 2)$ into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation, 

$\displaystyle y-2=\frac{2}{3}(x+1)$

$\displaystyle y-2=\frac{2}{3}x+\frac{2}{3}$

$\displaystyle y=\frac{2}{3}x+\frac{8}{3}$

For the second equation,

$\displaystyle y-2=-\frac{2}{3}(x+1)$

$\displaystyle y-2=-\frac{2}{3}x-\frac{2}{3}$

$\displaystyle y=-\frac{2}{3}x+\frac{4}{3}$

For a hyperbola with its foci on the $\displaystyle y$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the $\displaystyle x$ terms together and $\displaystyle y$ terms together.

$\displaystyle 4y^2-54x^2-24y+108x-234=0$

$\displaystyle 4y^2-24y-54x^2+108x-234=0$

Factor out $\displaystyle -54$ from the $\displaystyle x$ terms and $\displaystyle 4$ from the $\displaystyle y$ terms.

$\displaystyle 4(y^2-6y)-54(x^2-2x)-234=0$

Complete the squares. Remember to add the amount amount to both sides of the equation!

$\displaystyle 4(y^2-6y+9)-54(x^2-2x+1)-234=-18$

Add $\displaystyle 234$ to both sides of the equation:

$\displaystyle 4(y^2-6y+9)-54(x^2-2x+1)=216$

Divide both sides by $\displaystyle 216$.

$\displaystyle \frac{y^2-6y+9}{54}-\frac{x^2-2x+1}{4}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\displaystyle \frac{(y-3)^2}{54}-\frac{(x-1)^2}{4}=1$

The slopes of this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=2$ and $\displaystyle b=\sqrt{54}=3\sqrt6$.

Thus, the slopes for its asymptotes are $\displaystyle \pm\frac{3\sqrt6}{2}$.

Now, plug in the center of the hyperbola into the point-slope form of the equation of alien to get the equations for the asymptotes.

The center of the hyperbola is $\displaystyle (1, 3)$.

The equations for the asymptotes are then:

$\displaystyle y-3=\pm\frac{3\sqrt6}{2}(x-1)$

$\displaystyle y=\pm\frac{3\sqrt6}{2}(x-1)+3$

For a hyperbola with its foci on the $\displaystyle x$-axis, like the one given in the equation, recall the standard form of the equation:

$\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$, where $\displaystyle (h,k)$ is the center of the hyperbola.

The slopes of the asymptotes for this hyperbola are given by the following:

$\displaystyle m=\pm\frac{b}{a}$

For the hyperbola in question, $\displaystyle a=\sqrt{24}=2\sqrt6$ and $\displaystyle b=5$.

Thus, the slopes for its asymptotes are $\displaystyle m=\pm\frac{5}{2\sqrt6}=\pm\frac{5\sqrt6}{12}$.

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at $\displaystyle (5,-3)$.

To find the equations of the asymptotes, use the point-slope form of a line.

$\displaystyle y+3=\pm\frac{5\sqrt6}{12}(x-5)$

$\displaystyle y=\frac{5\sqrt6}{12}(x-5)-3$