Math / Geometry

QuestionQuestion Graph the conic $\frac{(y+6)^{2}}{9}-(x+6)^{2}=1$

Studdy Solution

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 \]

STEP 6

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.

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