Question 8 On the coordinate plane below, graph the hyperbola represented by the following equation. = 1 16 . Draw the auxiliary rectangle and the asymptotes. ● Draw and label the vertices, the center, and the two focal points.

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Chapter7: Conic Sections And Quadratic Systems
Section7.CT: Chapter Test
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**Question 8**

On the coordinate plane below, graph the hyperbola represented by the following equation.

\[
\frac{x^2}{9} - \frac{y^2}{16} = 1
\]

- Draw the auxiliary rectangle and the asymptotes.
- Draw and label the vertices, the center, and the two focal points.

The following is a description of the graph:

1. **Axes and Grid**: The graph is plotted on a rectangular coordinate system with the x-axis and y-axis intersecting at the origin (0,0). The grid has intervals indicating 1 unit per division on both axes.

2. **Hyperbola Equation**: The given equation is \(\frac{x^2}{9} - \frac{y^2}{16} = 1\), which is the standard form for a hyperbola centered at the origin with the transverse axis along the x-axis. 

3. **Vertices and Center**:
   - **Center (C)**: The center of the hyperbola is at the origin (0,0).
   - **Vertices (V1 and V2)**: The vertices are located at (\(\pm a, 0\)), where \(a = \sqrt{9} = 3\). Thus, the vertices are at \(V1 (3, 0)\) and \(V2 (-3, 0)\).

4. **Focal Points**:
   - The focal points (F1 and F2) lie along the transverse axis (x-axis) and can be found using the relationship \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to each focus.
   - Here, \(a^2 = 9\) and \(b^2 = 16\).
   - Thus, \(c = \sqrt{9 + 16} = \sqrt{25} = 5\).
   - The focal points are located at \(F1 (5, 0)\) and \(F2 (-5, 0)\).

5. **Auxiliary Rectangle and Asymptotes**:
   - The auxiliary rectangle helps in sketching the asymptotes.
   - The sides of this rectangle are parallel to the axes and pass through the vertices and co-vertices. The dimensions are \(2a\) along the x-axis and \(2b\) along
Transcribed Image Text:**Question 8** On the coordinate plane below, graph the hyperbola represented by the following equation. \[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \] - Draw the auxiliary rectangle and the asymptotes. - Draw and label the vertices, the center, and the two focal points. The following is a description of the graph: 1. **Axes and Grid**: The graph is plotted on a rectangular coordinate system with the x-axis and y-axis intersecting at the origin (0,0). The grid has intervals indicating 1 unit per division on both axes. 2. **Hyperbola Equation**: The given equation is \(\frac{x^2}{9} - \frac{y^2}{16} = 1\), which is the standard form for a hyperbola centered at the origin with the transverse axis along the x-axis. 3. **Vertices and Center**: - **Center (C)**: The center of the hyperbola is at the origin (0,0). - **Vertices (V1 and V2)**: The vertices are located at (\(\pm a, 0\)), where \(a = \sqrt{9} = 3\). Thus, the vertices are at \(V1 (3, 0)\) and \(V2 (-3, 0)\). 4. **Focal Points**: - The focal points (F1 and F2) lie along the transverse axis (x-axis) and can be found using the relationship \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to each focus. - Here, \(a^2 = 9\) and \(b^2 = 16\). - Thus, \(c = \sqrt{9 + 16} = \sqrt{25} = 5\). - The focal points are located at \(F1 (5, 0)\) and \(F2 (-5, 0)\). 5. **Auxiliary Rectangle and Asymptotes**: - The auxiliary rectangle helps in sketching the asymptotes. - The sides of this rectangle are parallel to the axes and pass through the vertices and co-vertices. The dimensions are \(2a\) along the x-axis and \(2b\) along
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