Math Snap

STEP 1

1. The given conic is in the form of a hyperbola.
2. The equation is centered at $(-6, -6)$.
3. The hyperbola opens vertically.

STEP 2

1. Identify the type of conic and its properties.
2. Determine the center of the hyperbola.
3. Identify the vertices and asymptotes.
4. Sketch the graph of the hyperbola.

STEP 3

Identify the type of conic and its properties:
The given equation is $\frac{(y+6)^{2}}{9} - (x+6)^{2} = 1$, which is in the standard form of a vertical hyperbola $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$.

STEP 4

Determine the center of the hyperbola:
The center $(h, k)$ is $(-6, -6)$.

STEP 5

Identify the vertices and asymptotes:
- For a vertical hyperbola, the vertices are located at $(h, k \pm a)$.
- Here, $a^2 = 9$, so $a = 3$.
- The vertices are $(-6, -6 \pm 3)$, which are $(-6, -3)$ and $(-6, -9)$.
- The asymptotes for a vertical hyperbola are given by the equations:
$$ y = k \pm \frac{a}{b}(x-h)
\] Here, $b^2 = 1$, so $b = 1$.
The asymptotes are:
$$ y = -6 \pm 3(x + 6)
\] Simplifying, the equations are:
$$ y = 3x + 12
\] $$ y = -3x - 24
\]

Sketch the graph of the hyperbola:
- Plot the center at $(-6, -6)$.
- Plot the vertices at $(-6, -3)$ and $(-6, -9)$.
- Draw the asymptotes using the equations $y = 3x + 12$ and $y = -3x - 24$.
- Sketch the branches of the hyperbola opening upwards and downwards, approaching the asymptotes.
The graph of the hyperbola is now complete.